MA442-Final Review

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Find the length of the curve defined by y=2ln⁡((x/2)^2−1) from x=5 to x=8.

3.67

Find the limit of the sequence: an=(3n^2+7n+3)/(2n^2+9n+9) Limit =

3/2

Evaluate the limit using L'Hospital's rule lim x→0 e^x+2x−1/2x

3/2

The function sqrt((x^2+3x+14)−x) has one horizontal asymptote at y=

3/2

Evaluate the indefinite integral. ∫9^2dx/(9+x^2)^2 =

3/2(arctan(x/3)+1/2sin(2arctan(x/3)))+C

Find the limit of the sequence an=(3n−4)/(8n+8)

3/8

Find the length of parametrized curve given by x(t)=6t^2+12t, y(t)=−4t^3−12t^2−9t where t goes from 0 to 1. Answer:

Find the length of the curve defined by y=3x^(3/2)−1 from x=4 to x=7.

31.70

Evaluate the definite integral. ∫π/15 0 sec^5⁡(5x)/cot⁡(5x) dx [NOTE: Remember to enter all necessary *, (, and ) !! Enter arctan(x) for tan−1⁡x , sin(x) for sin⁡x . ]

31/25

Use l'Hospital's Rule to evaluate the limit. lim x→0 sin⁡6x/sin⁡7x Limit:

6/7

Find the length of the curve defined by y=3x^(3/2)+5. from x=5 to x=10.

61.53

Rewrite the expression 5log⁡x−4log⁡(x^2+1)+4log⁡(x−1) as a single logarithm log⁡A. Then the function A=

(x^5(x-1)^4)/(x^2+1)^4

Find the integral. ∫e^(7x)sin⁡(6x)dx=

(−6e^(7x)cos(6x)/85)+(7e^(7x)sin(6x)/85)+C

∫4 2. ln⁡(2/x)/x. dx=

-(1/2)*ln(2)^(2)

Evaluate the definite integral. ∫14sin⁡(π/6) 0 x^3/sqrt(196−x^2) dx =

-1029(sqrt(3))+(5488/3)

Evaluate the integral ∫1 0 (−7x^3+28x+70)/(x^2−x−6)dx

-13.338

Compute the following limit using l'H\^opital's rule if appropriate. Use INF to denote ∞ and MINF to denote −∞. lim x→∞. (sqrt^3(x^3−7x^2)−x) =

-2.33333

Evaluate the integral ∫x^2dx/sqrt(36−x^2) using trigonometric substitution.

18(arcsin(1/6x)−(1/2)sin(2arcsin(1/6x)))+C

∫sqrt(12x−x^2) dx =

18(sin^−1((1/6)x−1)+(1/2)sin(2sinz−1((1/6)x−1)))+C

Find the length of the curve defined by y=2x^(3/2)−1 from x=3 to x=6.

19.24

Evaluate the integral. ∫3xsin⁡(−2x)dx Answer =

3(1/2xcos(2x)−1/4sin(2x))

∫e sqrt(e). 5/(tln⁡(t2)) dt=

(5/2)*ln(2)

Find the length of the curve defined by y=2x^(3/2)+7 from x=5 to x=9.

31.89

∫1/sqrt(25−81x^2) dx = WeBWorK notation for sin−1⁡(x) is arcsin(x) or asin(x), for tan−1⁡( x) it's arctan(x) or atan(x).

(1/9)arcsin((9/5)x) +C

Consider the following parametric equation. x=14(cos⁡θ+θsin⁡θ) y=14(sin⁡θ−θcos⁡θ) What is the length of the curve for θ=0 to θ=1/2π? Answer:

.5*14*(1*pi/2)^2

What is the length of the curve ln⁡(sec⁡(x)) from x=0 to x=1?

1.22619

What is the length of the curve ln⁡(sec⁡(x)) from x=0 to x=1?

1.23

Referring to the figure above, find the volume generated by rotating the region R2 about the line OA. volume =

1.5708

Find the length of the curve defined by y=2ln⁡((x/2)^2−1) from x=3 to x=4.

2.02

∫1 0. x^2 (^3sqrte^x) dx =

39(^3√e−54)

Find the volume of the solid that results when the red region is revolved about the x-axis. f(x)=sqrt(5−x), a=−1, b=5 volume =

Area: A(x) = πr(x)2 = π(5−x)2 = π(5−x) Volume: =∫5 -1. A(x)dx = π∫5 -1. [5−x]dx =π[5x−x22]5 -1 = π[(252)−(−112)] = 18π The volume of the solid revolved around the x-axis is 18π.

Find the common ratio and write out the first four terms of the geometric sequence {6^(n+1)/2} The common ratio is = a1= a2= a3= a4=

Common ratio = 6 a1= 18 a2= 108 a3= 648 a4= 3888

Suppose a particle moves along a straight line with velocity v(t)=t^2e^(−2t) meters per second after t seconds. It travels ____ meters during the first t seconds.

Since v(t)>0 for all t, the distance is s(t)=∫t 0 v(w)dw= ∫t 0 w^2e^(−2w) dw. First let u=w^2, dv=e^(−2w) dw⇒du=2w dw, v=−1/2e^(−2w). Then s(t)=[(−1/2)w^2e^(−2w)]t 0 +∫t 0 we^(−2w) dw= (−1/2)t^2e^(−2t)+∫t 0 we^(−2w) dw Next let U=w, dV= e^(−2w)dw⇒dU=dw, V=−1/2e^(−2w). Then s(t)=(−1/2)t^2e^(−2t)+[(−1/2)we^(−2w)]t 0 +∫t 0 1/2e^(−2w)dw= (−1/2)t^2e^(−2t)−(1/2)te^(−2t)+1/2∫t 0. e^(−2w)dw= (−1/2)t^2e^(−2t)−(1/2)te^(−2t)+1/2[−1/2e^(−2w)]t 0= (−1/2)t^2e^(−2t)−(1/2)te^(−2t)−1/4[e^(−2t)−1]

Calculate the Taylor polynomials T2(x) and T3(x) centered at x=1 for f(x)=ln(x+1). T2(x)= T3(x)=T2(x)+

T2(x)= 0.693147+0.5*(x-1)-0.125*(x-1)^2 T3(x)= 0.0416667*(x-1)^3

Compute T2(x) at x=0.2 for y=e^x and use a calculator to compute the error |e^x−T2(x)| at x=0.7. T2(x)= |e^x−T2(x)| =

T2(x)= e^0.2+e^0.2*(x-0.2)+e^0.2/2*(x-0.2)^2 |e^x−T2(x)| = 0.0289732

Find T4(x): the Taylor polynomial of degree 4 of the function f(x)=arctan⁡(14x) at a=0. (You need to enter a function.) T4(x)=

T4(x)= (14*x-(14^3)*x^3/3)

Find T5(x), the degree 5 Taylor polynomial of the function f(x)=cos⁡(x) at a=0. T5(x)= Find all values of x for which this approximation is within 0.001048 of the right answer. Assume for simplicity that we limit ourselves to |x|≤1. |x|≤

T5(x)= (1-x^2/2+x^4/24) |x|≤ 0.954

Find the value of ∫π/5 0. cos⁡(2x)dx. Remember: The angles for sin and cosine are always (well... almost always) in radians!

Using the substitution u=2x (and so du=2dx) we get: ∫π/5 0 cos⁡(2x)dx= (1/2)∫π/5 0 cos⁡(2x)⋅2dx= (1/2)∫π/5 0 cos⁡(u)du= (1/2)(sin⁡(2x))|π/5 0 = (1/2)(sin⁡(2π/5)−sin⁡(0))= (1/2)sin⁡(2π/5)

Evaluate the definite integral. ∫3 0 1/sqrt(81+x^2) dx =

ln((1/3)+(sqrt(10)/3))

Let f(x)=−3x+7. Find f^−1(x). f^−1(x)= . Now for fun, verify that (f∘f^−1)(x)=(f^−1∘f)(x)=x

(-x/3)+2.3333

Let f(x)=ln(sqrt(6x+6)(3x−4)) f′(x)=

(1/(sqrt(6x+6)(3x-4)))(1/2(sqrt(6x+6)(3x-4)))(6(3x-4)+3(6x+6))

∫ln⁡(7t+9)/(7t+9) dt=

(1/14)*ln(7*t + 9)^(2) + C

∫(ln⁡(sin⁡(x))cos⁡(x))/(sin⁡(x)) dx=

(1/2)*ln(sin(x))^(2) + C

∫e 0. ln⁡(2s+1)/(2s+1). ds=

(1/4)*ln(1 + 2*e)^(2)

∫4 1. ln⁡(y2)/y. dy=

(1/4)*ln(16)^(2)

Evaluate the indefinite integral. ∫(x^3)e^(x^4) dx Answer =

(1/4)e^(x^4)+C

Find the inverse function of f(x)=1/(x+5). f^−1(x)=

(1/x)-5

f(x)=1/x+6 f^−1(x)=

(1/x)-6

f(x)=13−x^2, x≥0 f^−1(x)=

(13-x)^0.5

Find f′(x) if f(x)=ln sqrt((⁡2x−4)/(9x+9)). f′(x)=

(2/(2x+(-4))-9/(9x+9))/2

Find the inverse function to y=f(x)=(4−5x)/(8−4x). x=g(y)=

(4-8y)/(5-4y)

Find the inverse function of f(x)=sqrt(8x+6). f^−1(x)=

(x^2-6)/8

The domain of the function g(x)=loga⁡(x^2−25) is (−∞,___ ) and ( ___ , ∞).

(−∞,-5 ) and ( 5 , ∞).

Find the volume of the solid that results when the region bounded by x=y^2 and x=y+12 is revolved about the y-axis. volume =

First we find the points of intersection y^2=y+12 (y−4)(y+3)=0 So y=−3 and y=4 are points of intersection (and hence our limits of integration). Area: A1(y) = πr2(y)^2 = π(y+12)^2 = π(y^2+24y+144) A2(y)=πr1(y)^2 =π (y^2)^2 =π(y^4) Volume =∫4 -3[A1(y)−A2(y)]dy = π∫4 -3 [y^2+24y+144−y^4]dy =π[y^3/3+12y^2+144y−y^5/5]4 -3 = π[(8768/15)−(−1422/5)] =13034/15π The volume of the solid revolved around the y-axis is 13034/15π.

Find the volume of the solid whose base is the semicircle y= sqrt (4−x^2) where −2 ≤ x ≤ 2, and the cross-sections perpendicular to the x-axis are squares. V =

For each x, the base of the square cross section extends from the semicircle y=sqrt(4−x^2) to the x-axis. The square therefore has a base with length sqrt(4−x^2) and an area of sqrt(4−x^2)^2= 4−x^2. The volume of the solid is then ∫2 -2 (4−x^2)dx =(4x − 1/3x^3)|2 -2 = 32/3 or 10.666667

Use cylindrical shells to find the volume of the solid generated when the region enclosed by the given curves is revolved about the y-axis. y=1/x,x=1,x=4,y=0 Volume =

For f(x)=1/x Volume =2π∫4 1. x⋅f(x)dx =2π∫4 1. 1dx =2π[x]4 1 =6π The volume of the solid revolved around the y-axis is 6π.

∫(ln⁡(2t+5))/(2t+5) dt=

The correct answer to this problem is (1/4)*ln(2*t + 5)^(2) + C

∫sqrt(ln⁡(4z))/z dz=

The correct answer to this problem is (2/3)*ln(4*z)^(3/2) + C by using substitution.

∫(sin⁡(ln⁡(1/x)))/x dx=

The correct answer to this problem is cos(ln(1/x)) + C by using substitution.

∫(sin⁡(ln⁡(y2)))/y dy=

The correct answer to this problem is - (1/2)*cos(ln(y^(2))) + C by using substitution.

Find the volume of the solid generated when the region enclosed by y=sqrt(x+4) , y=sqrt(3x) and y=0 is revolved about the x-axis. [Hint: Split the solid into two parts.] volume =

The curves intersect when x=4. The first segment goes from −4 to 0 with area: A1(x) = πr1(x)2 = π(sqrt(x+4))^2 = π(x+4) The second segment goes from 0 to 2 with area: A2(x) = π[r1(x)^2−r2(x)^2] = π[(sqrt(x+4))^2−(3x)^2] = π(−2x+4) Volume = π∫0 -4[x+4]dx+π∫2 0[−2x+4]dx = π[x^2/2+4x]0 -4+π[4x−1x2]2 . 0 = 8π The volume of the solid revolved around the x-axis is 8π.

Find the volume of the solid that results when the red region is revolved about the y-axis. f(x)=x^2−1,b=4,(b,f(b))=(4,15) volume =

The limits of integration will be from y=0 to y=f(b)=f(4)=15 and the area will be the difference between the areas from x=0 to x=4 and x=0 to x=sqrt(y+1). Area: A(y)=πr1(y)^2−πr2(y)^2 =π[(4)^2−(sqrt(y+1))^2] =π(15−y) Volume =∫15 0A(y)dy =π∫15 0[15−y]dy =π[15y−y^2/2]15 0 =225/2π The volume of the solid revolved around the y-axis is 225/2π.

Find dy/dx for each of the following functions y=ln⁡(6x−11/x^7sqrt(x2+1)) dy/dx= y=x^cos⁡(x) dy/dx=

dy/dx = (6/6x-11)-(1/x)-(2x/7(x^2+1)) dy/dx = x^cos(x)((cos(x)/x)-sin(x)ln(x))

Let f(x)=sqrt(sin⁡(e^((x^3)cos⁡(x)))) f′(x)=

e^((x^3)cosx)cos(e^((x^3)cosx))(3x^2*cos(x)-X^3*sin(x))/2sqrt(sin(e^((x^3)cosx))

Assume that the function f is a one-to-one function. (a) If f(9)=8, find f^−1(8). Your answer is (b) If f^−1(−4)=−5, find f(−5). Your answer is

f^−1(8) = 9 f(−5) = -4

If f(x)=14x−13, then f^−1(y)= f^−1(−7)=

f^−1(y) = (y+13)/14 f^−1(−7) = 0.4286

If f(x)=2cos(2ln(x)), find f′(x). Find f′(1).

f′(x) = -2*2(1/x)sin(2ln(x)) f′(1) = 0

Let f(x)=−3x^2ln⁡x f′(x)= f′(e^4)=

f′(x) = -3x^(2-1)(2ln(x)+1) f′(e^4) = -3e(4(2-1))(2(4)+1)

Let f(x)=(ln⁡x)^6 f′(x)= f′(e^2)=

f′(x) = 6(ln⁡x)^5d/dx(ln⁡x) = 6(ln⁡x)^5/x f′(e2) = 6(ln⁡(e^2)^5*1/e^2 = 6⋅2^5⋅1/e^2 = 192/e^2

If f(x)=8ln⁡(8x+4ln⁡(x)), find f′(x). f′(x)=

f′(x)= (8/8x+4ln(x))(8+(4/x))

Suppose f(x)=ln⁡(ex^6/(x−4)^4). (a) Find f′(x)=. (Hint: Apply the laws of logarithms to f(x) before taking its derivative.) (b) Find d/dx(e^f(x))=

f′(x)= 2(x−12)/x(x−4) d/dx(e^f(x))= 2ex^5(x−12)/(x−4)^5

If f(x)=3ln(3+x), find f′(x). f′(x)= Find f′(4). f′(4)=

f′(x)= 3(1/3+x) f′(4)= 0.428571

Let f(x)=x^2(x−7)^6/(x^2+6)^2 Use logarithmic differentiation to determine the derivative. f′(x)= f′(7)=

f′(x)= x^2(x-7)^6/(x^2+6)^2((2/x)+(6/x-7)-(4x/x^2+6)) f′(7)= 0

Let f(x)=10(sin(x))^x. Find f′(2). f′(2)=

−8.3542

Evaluate the indefinite integral. ∫9e^(9x)sin(e^9x) dx

−cos(e^9x)+C

If f(x)=3x−4 and g(x)=(x+4)/3, (a) f(g(x))= (b) g(f(x))= (c) Thus g(x) is called an _____ function of f(x)

(a) x (b) x (c) inverse

Evaluate the definite integral. ∫π/3 0. e^(sin⁡(x))cos⁡(x) dx

(e^sqrt(3/2))−1

Evaluate the indefinite integral. ∫tan^5⁡xsec^5⁡xdx Answer:

(sec^9(x)/9)−(2sec^7(x)/7)+(sec^5(x)/5) + C

Let f(x)=2+sqrt(x−2). Find f^−1(x). f^−1(x)=

(x-2)^2+2

Find the inverse function of f(x)=9+3sqrt(x). f^−1(x)=

(x-9)^3

Let y=tan^−1⁡(sqrt(8x^2−1)),then dy/dx=

1/(xsqrt(8x^2-1))

Use integration by parts to evaluate the definite integral. ∫e 1 2t^2ln⁡t dt

2(2e^3+1)/9

Evaluate the limit lim x→∞. sqrt(x^2+5x+12)−x

5/2

Evaluate the indefinite integral ∫dx/(x^2+11x+30) =

ln|x+5|−ln|x+6|+C

∫π/3 0 sin4⁡(3x)dx =

π/8

Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If it diverges to infinity, state your answer as INF. If it diverges to negative infinity, state your answer as MINF. If it diverges without being infinity or negative infinity, state your answer as DIV. ∫10 2 4/(x−3)^2dx

∞

Determine whether the sequence is divergent or convergent. If it is convergent, evaluate its limit. If it diverges to infinity, state your answer as INF. If it diverges to negative infinity, state your answer as MINF. If it diverges without being infinity or negative infinity, state your answer as DIV. lim n→∞ 9n+(19/12^n)

∞

Find the volume of the solid that results when the region bounded by y=sqrt(169−x^2) and y=5 is revolved about the x-axis. volume -

Area: A1(x)= πr(x)^2 = π(sqrt(169−x^2))^2 = π(sqrt(169−x^2)) A2(x)= πr(x)^2 = π(5)^2 = 25π Volume =∫12 -12. [A1(x−A2(x)]dx = π∫12 -12. [144−x^2]dx = 2π∫12 0 [144−x^2]dx = π[144x−x33]12 0 = 2π[(1152)−(0)] =. 2304π The volume of the solid revolved around the x-axis is 2304π.

Find the largest interval containing x=0 that the Remainder Estimation Theorem allows over which f(x)=sin(3x) can be approximated by p(x)=3x−((3x)^3/6) to three decimal-place accuracy throughout the interval. Check your answer by graphing |f(x)−p(x)| over the interval you obtained. Enter Interval in Interval Notation.

(-0.1899,0.1899)

For the following indefinite integral, find the full power series centered at x=0 and then give the first 5 nonzero terms of the power series and the open interval of convergence. f(x)=∫x2ln⁡(1+x) dx ∞ f(x)=C+∑ n=1 f(x)=C+__+ ___+ ___+ ___+ ___+⋯ The open interval of convergence is: (Give your answer in interval notation.)

(-1)^(n+1)*x^(2+n+1)/[n*(2+n+1)] f(x)= C + x^(2+2)/[1*(2+2)] + -1*x^(2+3)/[2*(2+3)] + x^(2+4)/[3*(2+4)] + -1*x^(2+5)/[4*(2+5)] + x^(2+6)/[5*(2+6)] (-1,1)

For the following indefinite integral, find the full power series centered at x=0 and then give the first 5 nonzero terms of the power series and the open interval of convergence. f(t)=∫tan^−1⁡(6x)/x dx ∞ f(t)=C+∑ n=0 f(t)=C+__+ ___+ ___+ ___+ ___+⋯ The open interval of convergence is: (Give your answer in interval notation.)

(-1)^n*6^(2*n+1)*x^(2*n+1)/[(2*n+1)^2] f(t)= C + 6^1*x^1/(1^2) + -1*6^3*x^3/(3^2) + 6^5*x^5/(5^2) + -1*6^7*x^7/(7^2) + 6^9*x^9/(9^2) (-0.166667,0.166667)

For the following function, find the full power series centered at x=0 and then give the first 5 nonzero terms of the power series and the open interval of convergence. f(x)=x^3/(1+6x)^2 ∞ f(x)=∑ n=0 f(x)= __+ ___+ ___+ ___+ ___+⋯ The open interval of convergence is: (Give your answer in interval notation.)

(-1)^n*x^(n+3)*(n+1)*6^n f(x)= x^(0+3) + -1*x^(1+3)*2*6 + x^(2+3)*3*6^2 + -1*x^(3+3)*4*6^3 + x^(4+3)*5*6^4 (-0.166667,0.166667)

For what values of p is the series convergent? ∞ ∑ (−1)^(n−1)((ln⁡n)^p/n^4) n=2 Answer (in interval notation):

(-∞,∞)

Given the equation: xy=7, set up an integral to find the length of path from x=a to x=b and enter the integrand below. (i.e. if your integral is L=∫b a. c^2x^2/h dx enter c^2x^2/h as your answer.) L=∫b a __ dx

(1+7^2/x^4)^.5

Evaluate the indefinite integral. ∫sin^3⁡(11x)cos^4⁡(11x)dx

(1/11)((−cos^5(11x)/5)+(cos^7(11x)/7)) +C

Evaluate the indefinite integral. ∫sec^6⁡xtan^(12)⁡x dx=

(1/17)tan^(17)(x)+(2/15)tan^(15)(x)+(1/13)tan^(13)(x) + C

Evaluate the indefinite integral. ∫sin⁡(8x)sin⁡(17x) dx

(1/2)((1/9)sin(9x)−(1/25)sin(25x)) +C

Evaluate the indefinite integral. ∫sin⁡(4x)cos⁡(6x) dx=

(1/2)(−(1/10)cos(10x)+(1/2)cos(2x)) +C

Evaluate the indefinite integral. ∫sec^3⁡x dx Answer:

(1/2)sec(x)tan(x)+(1/2)ln|tan(x)+sec(x)| + C

Evaluate the indefinite integral. ∫xarctan⁡(7x)dx Answer:

(1/2)x^2arctan(7x)−(1/98)(−arctan(7x)+7x) + C

Evaluate the indefinite integral. ∫tan^4⁡x dx Answer:

(1/3)tan^3(x)+x−tan(x) + C

Evaluate the indefinite integral. ∫x^2arctan⁡(3x)dx Answer:

(1/3)x^3arctan(3x)−(1/162)(9x^2+1−ln|9x^2+1|) + C

Evaluate the definite integral. ∫7 1 sin^2⁡(8x)cos^2⁡(8x) dx

(1/8)(6−(1/32)(sin(224)−sin(32)))

Evaluate the integral. ∫4 −1 1/(x+9)(x^2+16) dx Answer:

(1/97)(ln(13)−3ln(2))+(1/388)(−ln(1024/289)+9(π/4+arctan(1/4)))

Compute the following limits using L'Hospital's rule if appropriate. Use INF to denote ∞ and MINF to denote −∞. lim x→1. 5^x−5/x^2−1 = lim x→∞. tan^−1⁡(x)/(1/x)−5 =

(5/2)ln(5) −π/10

Find the arc length of the curve y=1/8(−x^2+8ln⁡(x)) from x=3 to x=8. Length =

(8*ln(8/3)/8)+((8^2-3^2)/8)

(a) Find the local quadratic approximation of sqrt(x/4) at x0=4. Local Quadratic Approx Formula = (b) Use the result obtained in part (a) to approximate sqrt(1.04) and compare your approximation to that produced directly by your calculating utility. sqrt(1.04)≈

(a) Let f(x)=sqrt(x/4), f′(x)=1/2sqrt(4x), f″(x)=−1/sqrt((4x)^3). f(x0)=1, f′(x0)=1/8, f″(x0)=−1/64 Local Quadratic Approximation: f(x)=sqrt(x/4)≈f(x0)+f′(x0)(x−x0)+(f″(x0)/2)(x−x0)^2= 1+(1/8)(x−4)−(1/128)(x−4)^2 (b) sqrt(1.04)=sqrt(4.16/4)=f(x0+0.16)≈ 1+1/8(0.16)−1/128(0.16)^2 = 1.0198 As a comparison numerical we have sqrt(1.04) = 1.0198

Evaluate the definite integral. ∫2 0 te^−t dt=

(−3/e^2)+1

∫dx/x^2sqrt(36−x^2) =

-((sqrt(36-x^2))/36x)+c

Let f(x) be a function that is defined and has a continuous derivative on the interval (2,∞). Assume also that f(4)=11 , |f(x)|<x^7+7 and ∫∞ 4 f(x)e^(−x/2)dx = −8 Determine the value of ∫∞ 4 f′(x)e^(−x/2) dx

-5.50

Use the partial fractions method to express the function as a power series (centered at x=0) and then give the open interval of convergence. f(x)=4x+5/(5x^2−19x−4) ∞ f(x)=∑ n=0 The open interval of convergence is: (Give your answer in interval notation.)

-[(-5)^n+1/[4^(n+1)]]x^n (-0.2,0.2)

Assume the world population will continue to grow exponentially with a growth constant k=0.0132 (corresponding to a doubling time of about 52 years), it takes 12 acre of land to supply food for one person, and there are 13,500,000 square miles of arable land in in the world. How long will it be before the world reaches the maximum population? Note: There were 6.06 billion people in the year 2000 and 1 square mile is 640 acres. Answer: The maximum population will be reached some time in the year = Hint: Convert .5 acres of land per person (for food) to the number of square miles needed per person. Use this and the number of arable square miles to get the maximum number of people which could exist on Earth. Proceed as you have in previous problems involving exponential growth.

.5 acre ×1sq mi/640acres=1/1280sq mi is the number of square miles needed per person. Since 13,500,000 square miles of arable land exist on earth we see that 17.28 billion is the maximum number of people that can exist on Earth. We plug this into the exponential growth equation to see, 17.28×10^9=6.06×10^9e^0.0132t. Using logs, we solve for t=2079.

Compute the following limit using l'Hopital's rule if appropriate. Use INF to denote ∞ and MINF to denote −∞. lim x→0+. 2sin⁡(x)ln⁡(x) =

Compute the value of the following improper integral. If it is divergent, type "Diverges" or "D". ∫∞ −∞ x^3e^(−x^4)dx Answer:

Determine whether the sequence is divergent or convergent. If it is convergent, evaluate its limit. (If it diverges to infinity, state your answer as inf . If it diverges to negative infinity, state your answer as -inf . If it diverges without being infinity or negative infinity, state your answer as div ) lim n→∞ (30+26arctan⁡(n!))/6^n Answer:

Determine whether the sequence is divergent or convergent. If it is convergent, evaluate its limit.(If it diverges to infinity, state your answer as inf . If it diverges to negative infinity, state your answer as -inf . If it diverges without being infinity or negative infinity, state your answer as div .) lim n→∞ 4(2^n)+18/16(5^n) Answer:

Evaluate the limit using L'Hopital's rule. lim x→∞ 5x^3/e^(7x)=

Find the limit of the sequence an=(cos⁡n)/4^n.

Find the following limits, using l'H\^opital's rule if appropriate lim x→∞. arctan⁡(x^4)/x^5 = lim x→0+. ^5sqrt(x)ln⁡(x) =

0 0

0.11019

Let T6(x): be the taylor polynomial of degree 6 of the function f(x)=ln⁡(1+x) at a=0. Suppose you approximate f(x) by T6(x), find all positive values of x for which this approximation is within 0.001 of the right answer. (Hint: use the alternating series approximation.) 0<x≤

0.492

What is the length of the curve ln⁡(sec⁡(x)) from x=0 to x=0.6?

0.6396

If a sequence c1,c2,c3,... has limit K then the sequence e^c1, e^c2, e^c3,... has limit eK. Use this fact together with l'Hopital's rule to compute the limit of the sequence given by bn=(n)^(1/n).

Let Tk(x) be the degree k Taylor polynomial of the function f(x)=sin⁡(x) at a=0. Suppose you approximate f(x) by Tk(x). If |x|≤1, how many terms are needed (that is, what is k) to obtain an error less than 1/6 ? Hint: use the alternating series approximation. Answer:

Determine whether the sequences are increasing, decreasing, or not monotonic. If increasing, enter 1 as your answer. If decreasing, enter −1 as your answer. If not monotonic, enter 0 as your answer. 1. an=cos⁡n/4^n 2. an=n−4/n+4 3. an=1/4n+6 4. an=sqrt(n+4)/6n+4

1. 0 2. 1 3. -1 4. -1

For each of the following forms determine whether the following limit type is indeterminate, always has a fixed finite value, or never has a fixed finite value. In the first case answer IND, in the second case enter the numerical value, and in the third case answer DNE. To discourage blind guessing, this problem is graded on the following scale 0-9 correct = 0 10-13 correct = .3 14-16 correct = .5 17-19 correct = .7 Note that l'Hospital's rule (in some form) may ONLY be applied to indeterminate forms. 1. 1^0 2. π^−∞ 3. ∞^−e 4. 1⋅∞ 5. ∞⋅∞ 6. ∞^∞ 7. ∞^1 8. ∞−∞ 9. 0^0 10. 1^−∞ 11. 1/−∞ 2. π^∞ 13. ∞/0 14. ∞^0 15. 0⋅∞ 16. 0^∞ 17. 0/∞ 18. 0^−∞ 19. ∞^−∞ 20. 1^∞

1. 1 2. 0 3. 0 4. dne 5. dne 6. dne 7. dne 8. ind 9. ind 10. ind 11. 0 12. dne 13. dne 14. ind 15. ind 16. 0 17. 0 18. dne 19. 0 20. ind

For each of the improper integrals below, if the comparison test applies, enter either A or B followed by one letter from C to K that best applies, and if the comparison test does not apply, enter only L. For example, one possible answer is BF, and another one is L. Hint: 0<e−x≤1 for x≥1. 1. ∫∞ 1 e^−x/x^2dx 2. ∫∞ 1 cos^2⁡(x)/(x^2+6)dx 3. ∫∞ 1 9+sin⁡(x)/sqrt(x−0.5)dx 4. ∫∞ 1 x/sqrt(x^6+6)dx 5. ∫∞ 1 1/(x^2+6)dx A. The integral is convergent B. The integral is divergent C. by comparison to ∫∞ 1 1/(x^2−6)dx. D. by comparison to ∫∞ 1 1/(x^2+6)dx. E. by comparison to ∫∞ 1 cos^2⁡(x)/x^2dx. F. by comparison to ∫∞ 1 e^x/x^2dx. G. by comparison to ∫∞ 1. −e^−x/2xdx. H. by comparison to ∫∞ 1 1/sqrt(x)dx. I. by comparison to ∫∞ 1 1/sqrt(x^5)dx. J. by comparison to ∫∞ 1 1/x^2dx. K. by comparison to ∫∞ 1 1/x^3dx. L. The comparison test does not apply.

1. AJ 2. AJ 3. BH 4. AJ 5. AJ

For each of the following series, tell whether or not you can apply the 3-condition test (i.e. the alternating series test). Enter D if the series diverges by this test, C if the series converges by this test, and N if you cannot apply this test (even if you know how the series behaves by some other test). 1. ∞ ∑ (−1)^n(n^10+1)/e^n n=1 2. ∞ ∑ (−1)^ncos⁡(n)/n^2 n=1 3. ∞ ∑ (−1)^n/n^5 n=1 4. ∞ ∑ (−1)^n(n^4+2n)/n^3−1 n=2 5. ∞ ∑ (−1)^n(n^3+1)/n^4+1 n=1 6. ∞ ∑ (−1)^n(n^3+1)/n^3+7 n=1

1. C 2. N 3. C 4. N 5. C 6. N

Match the functions with their graphs. Enter the letter of the graph below which corresponds to the function. 1. f(x)=ln⁡(−x) 2. f(x)=ln⁡(2−x) 3. f(x)=−ln⁡x 4. f(x)=2+ln⁡x 5. f(x)=−ln⁡(−x)

1. D 2. B 3. A 4. E 5. C

Consider the integral ∫(x^(21)−3x^(14)+12x^7−32)/(x^3−5x^2+4x)^3(x^4−256)^2dx Enter a T or an F in each answer space below to indicate whether or not a term of the given type occurs in the general form of the complete partial fractions decomposition of the integrand. A1,A2,A3... and B1,B2,B3,... denote constants. You must get all of the answers correct to receive credit. 1. A2x+B2/(x^4+256)^2 2. B1/x−4 3. A7x+B7/(x+4)^2 4. B4/(x−1)^4 5. A1x2 (a product, not a fraction!) 6. B5/x^4 7. B6/x+4 8. A3x+B3/x^2+16

1. F 2. T 3. F 4. F 5. T 6. F 7. T 8. T

Match each sequence below to the statement that BEST fits it. STATEMENTS Z. The sequence converges to zero; I. The sequence diverges to infinity; F. The sequence has a finite non-zero limit; D. The sequence diverges. SEQUENCES 1. n!/n^1000 2. (ln⁡(n))/n 3. n^100/(1.01)^n 4. sin⁡(n) 5. n^3−5n/3n−n^5 6. ln⁡(ln⁡(ln⁡(n))) 7. arctan⁡(n+1) 8. nsin(1/n)

1. I 2. Z 3. Z 4. D 5. Z 6. I 7. F 8. F

Determine whether each of the following integrals is proper, improper and convergent, or improper and divergent. 1. ∫9 0 1/sqrt^3(x−4)dx 2. ∫∞ 1 se^(−8s^2)ds 3. ∫∞ −∞ sin⁡(8z)dz 4. ∫25π −11π sin⁡(θ)arctan⁡(θ)dθ 5. ∫∞ 8 1/sqrt(t^2−64)dt 6. ∫∞ −∞ t/(t^2+11)dt 7. ∫9 4 ln⁡(x−4)dx 8. ∫9π/2 −π/4 tan^2⁡(8x)dx

1. Improper and convergent 2. Improper and convergent 3. Improper and divergent 4. Proper 5. Improper and divergent 6. Improper and divergent 7. Improper and convergent 8. Improper and divergent

Determine whether each of the following integrals is proper, improper and convergent, or improper and divergent. 1. ∫∞ −∞ t/t^2+10dt 2. ∫∞ 6 1/sqrt(t^2−36)dt 3. ∫∞ 1 se^(−6s^2)ds 4. ∫19 0 1/sqrt^3(x−9)dx 5. ∫19 9 ln⁡(x−9)dx 6. ∫∞ −∞ sin⁡(6z)dz 7. ∫19π/2 −π/9 tan^2⁡(6x)dx 8. ∫42π −10π sin⁡(θ)arctan⁡(θ)dθ

1. Improper and divergent 2. Improper and divergent 3. Improper and convergent 4. Improper and convergent 5. Improper and convergent 6. Improper and divergent 7. Improper and divergent 8. Proper

Determine whether each of the following integrals is proper, improper and convergent, or improper and divergent. 1. ∫∞ −∞ cos⁡(4x)dx 2. ∫6 2 1/(x−3)^3dx 3. ∫∞ 0 x^3e^(x^4)dx 4. ∫∞ 5 1/sqrt^3(x^2−4)dx 5. ∫∞ 0 arctan⁡(x)/(1+x^2)dx 6. ∫∞ 2 1/sqrt(x^3+4)dx 7. ∫0 −∞ x^2e^(x^3)dx 8. ∫∞ −∞ x/(x^2+10)dx

1. Improper and divergent 2. Improper and divergent 3. Improper and divergent 4. Improper and divergent 5. Improper and convergent 6. Improper and convergent 7. Improper and convergent 8. Improper and divergent

Enter a T or an F in each answer space below to indicate whether the corresponding statement is true or false. You must get all of the answers correct to receive credit. 1. ln⁡a^b=bln⁡a 2. log2⁡ab=log2⁡a+log2⁡b 3. ln⁡x/ln⁡y=ln⁡x−ln⁡y 4. ln⁡(x−y)=ln⁡x−ln⁡y

1. T 2. T 3. F 4. F

For each of the indefinite integrals below, choose which of the following substitutions would be most helpful in evaluating the integral. Enter the appropriate letter (A,B, or C) in each blank. DO NOT EVALUATE THE INTEGRALS. ] A. x=7tan⁡θ B. x=7sin⁡θ C. x=7sec⁡θ 1. ∫x249+x2dx 2. ∫x2−49dx 3. ∫x2dx49−x2 4. ∫(x2−49)5/2dx 5. ∫dx(49−x2)3/2

1. The integral ∫x^2sqrt(49+x^2) dx requires the substitution x=tan⁡θ because sqrt(7^2+x^2) is present in it. 2. The integral ∫sqrt(x^2−49) dx requires the substitution x=sec⁡θ because sqrt(x^2−7^2) is present in it. 3. The integral ∫x^2/sqrt(49−x^2) dx requires the substitution x=sin⁡θ because sqrt(7^2−x^2) is present in it. 4. The integral ∫(x^2−49)^(5/2) dx requires the substitution x=sec⁡θ because sqrt(7^2+x^2) is present in it, if you consider that ∫(x^2−49)^(5/2) dx = ∫(sqrt(x^2−7^2))^5 dx. 5.The integral ∫dx/(49−x2)^(3/2) requires the substitution x=sin⁡θ because sqrt(7^2+x^2) is present in it, if you consider that ∫dx/(49−x^2)(3/2) = ∫dx/(sqrt(7^2−x^2))^3.

Match each of the trigonometric expressions below with the equivalent non-trigonometric function from the following list. Enter the appropriate letter (A,B,C,D, or E) in each blank. A. tan⁡(θ) where x=7sin⁡θ B. cos⁡(θ) where x=7sin⁡θ C. (1/2)sin⁡(2θ) where x=7sin⁡θ D. sin⁡(θ) where x=7tan⁡θ E. cos⁡(θ) where x=7tan⁡θ 1. x4949−x2 = 2. x49−x2 = 3. x49+x2 = 4. 749+x2 = 5. 49−x27 =

1. Using the substitution x=7sin⁡θ and its inverse θ=sin^−1⁡(x/7), we get (x/49)sqrt(49−x^2) = (7sin⁡θ/7^2)sqrt(72−72sin2⁡θ = sin⁡θsqrt(7^2cos^2θ/7) = sin⁡θcos⁡θ = 1/2sin⁡2θ = 1/2sin⁡(2sin^−1⁡(x/7)) 2. Using the substitution x=7sin⁡θ and its inverse θ=sin^−1⁡(x/7), we get x/sqrt(49−x^2)= 7sin⁡θ/sqrt(7^2−7^2sin^2⁡θ)= 7sin⁡θ/(7^2cos^2⁡θ)= 7sin⁡θ/7cos⁡θ= tan⁡θ= tan⁡(sin−1⁡(x7)) 3. Using the substitution x=7tan⁡θ and its inverse θ=tan^−1⁡(x/7), we get x/sqrt(49+x^2)= 7tan⁡θ/sqrt(7^2+7^2tan^2⁡θ)= 7tan⁡θ/sqrt(7^2sec^2⁡θ)= 7tan⁡θ/7sec⁡θ= sin⁡θ= sin⁡(tan−1⁡(x7)) 4. Using the substitution x=7tan⁡θ and its inverse θ=tan^−1⁡(x/7), we get 7/sqrt(49+x^2)= 7/sqrt(7^2+7^2tan^2⁡θ)= 7/sqrt(7^2sec^2⁡θ)= 7/7sec⁡θ= cos⁡θ= cos⁡(tan^−1⁡(x/7)) 5. Using the substitution x=7sin⁡θ and its inverse θ=sin^−1⁡(x/7), we get 49−x^2/7= sqrt(7^2−7^2sin^2⁡θ)/7= sqrt(7^2cos^2⁡θ)/7= 7cos⁡θ/7= cos⁡θ= cos⁡(sin^−1⁡(x/7))

Match each sequence below to the statement that BEST fits it. STATEMENTS Z. The sequence converges to zero; I. The sequence diverges to positive infinity; F. The sequence has a finite non-zero limit; D. The sequence diverges, but not to infinity. SEQUENCES 1. 100n^2+1/3n! 2. (−5)^n/n! 3. (−1)^−n(2n/ln⁡(n)) 4. (5n^2n)^(1/n) 5. 5^n/n! 6. (e/10)^n 7. cos^2⁡(n)+sin^2⁡(n) 8. sqrt(n^2+4n)−sqrt(n^2)

1. Z 2. Z 3. D 4. I 5. Z 6. Z 7. F 8. F

Use an appropriate local quadratic approximation to approximate tan⁡60.8∘, and compare the result to that produced directly by your calculating utility. Enter the local quadratic approximation of tan⁡60.8∘. tan⁡60.8∘≈

1.79

Find the value of ∫π/4 0 sin⁡(2x)sin⁡(x)dx.

1/(3√2)

Find the limit of the sequence whose terms are given by an=(1/(e^(4n)+n^2))^(1/n).

1/(e^4)

Evaluate the limit using L'Hospital's rule if necessary lim x→0 e^x−1/sin⁡(11x)

1/11

Use integration by parts to evaluate the integral. ∫xe^4x dx=

1/16(e^(4x)·4x−e^(4x)) + C

Evaluate the following limit. Enter -I if your answer is −∞, enter I if your answer is ∞, and enter DNE if the limit does not exist. lim x→0. (1/7x−1/e^(7x)−1)=

1/2

Evaluate the indefinite integral. ∫x^3+52/(x^2+6x+8)dx =

1/2(x+3)^2+14ln|x^2+6x+8|−16(1/2ln|x+4|−1/2ln|x+2|)−9(x+3) +C

Evaluate the indefinite integral. ∫xcos^2⁡(2x)dx

1/32(8x^2+4xsin(4x)+cos(4x))

Evaluate the integral ∫x^3sqrt(4−x^2)dx using trigonometric substitution.

1/5(4-x^2)^(5/2)-4/3(4-x^2)^(3/2)+c

Evaluate the indefinite integral. ∫x^(13)cos(x^7)dx = Hint: First make a substitution and then use integration by parts to evaluate the integral.

1/7(x^7sin(x^7)+cos(x^7)) +C.

Use integration by parts to evaluate the integral. ∫2xcos⁡(4x)dx=

1/8(4xsin(4x)+cos(4x)) + C

Evaluate the following limit: lim x→∞. (18x/18x+6)^4x Enter -I if your answer is −∞, enter I if your answer is ∞, and enter DNE if the limit does not exist. Limit =

1/e^(4/3)

Compute the following limit using l'H\^opital's rule if appropriate. Use INF to denote ∞ and MINF to denote −∞. lim x→∞. (1−8/x)^x =

1/e^8

Consider the following parametric equation. x=16(cos⁡θ+θsin⁡θ) y=16(sin⁡θ−θcos⁡θ) What is the length of the curve for θ=0 to θ=(7/6)π? Answer:

107.47

How many terms of the series do we need to add in order to find the sum to the indicated accuracy? ∞ ∑ (−1)^(n−1)/n^2, error≤0.007. n=1 Answer:

Evaluate the indefinite integral. ∫84cos^3⁡(7x)dx

12(sin(7x)−(1/3)sin^3(7x)) +C

Evaluate the indefinite integral. ∫156cos^4⁡(12x)dx

13(cos^3(12x)sin(12x)+(3/8)(12x−(1/4)sin(48x)))

Consider the following parametric equation. x=5(cos⁡θ+θsin⁡θ) y=5(sin⁡θ−θcos⁡θ) What is the length of the curve for θ=0 to θ=(3/4)π? Answer:

13.88

Evaluate the following limit using L'Hospital's rule where appropriate. lim x→0 sin⁡(13x)/tan⁡(5x) Answer:

13/5

Consider the parametric curve given by the equations x(t)=t^2+7t+18 y(t)=t^2+7t+24 How many units of distance are covered by the point P(t)=(x(t),y(t)) between t=0 and t=7 ? Answer:

138.59

Let f(x)=7sin^−1(x^2) f′(x)= NOTE: The webwork system will accept arcsin⁡(x) or sin−1⁡(x) as the inverse of sin⁡(x).

14x/(√1−x^4)

Find the length of the given curve. x=3y^(4/3)−(3y^(2/3)/32), −64≤y≤64 L =

1539

Find the length of a parametrized curve given by x(t)=6t^2+6t, y(t)=−4t^3−6t^2 where t goes from 0 to 1. Answer:

Evaluate the integral ∫sqrt(64+16x^2)dx using trigonometric substitution. Note: Use C for an arbitrary constant, and be sure to absorb as much into C as is possible.

16(1/8x(sqrt(x^2+4))+1/2ln(abs(x+(sqrt(4+x^2)))/2))+c

Let f(x)=4sin^−1(x^4) f′(x)= NOTE: The webwork system will accept arcsin⁡(x) or sin−1⁡(x) as the inverse of sin⁡(x).

16x^3/(√1−x^8)

Consider the parametric curve given by the equations x(t)=t^2+12t−20 y(t)=t^2+12t+35 How many units of distance are covered by the point P(t)=(x(t),y(t)) between t=0 and t=7 ? Answer:

188.09

Evaluate the indefinite integral. ∫4xsec^6⁡(x^2)dx

2(tan(x^2)+(2/3)tan^3(x^2)+(1/5)tan^5(x^2)) +C

Find the limit of the sequence whose terms are given by an=(n^2)(1−cos⁡(2.1/n)).

2.205

Let T6(x) be the Taylor polynomial of degree 6 for the function f(x)=cos⁡(x) at a=0. Suppose you approximate f(x) by T6(x). If |x|≤1, what is the bound for your error of your estimate? Hint: use the alternating series approximation. Answer:

2.48*10^-5

If f(x)=5arctan⁡(4e^x), find f′(x)=

20e^x/16e^(2x)+1

Sam and Mary want to purchase a house. Suppose they invest 500 dollars into a mutual fund at the end of each month. How much will they have for a downpayment after 3 years if the per annum rate of return of the mutual fund is assumed to be 11percent compounded monthly?

21211.56

Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If not, give the answer -1. ∫∞ 7 xe^(−3x)dx

22/9e^(21)

Evaluate the integral. ∫2 −3 x^3+4/(x+9)(x+9) dx Answer:

243(ln(11)−ln(6))−(4865/33)

Let L be the circle in the x-y plane with the center of the origin and radius 76. Let S be a moveable circle with a radius of 24. S is rolled along the inside of L without slipping while L remains fixed. A point P is marked on S before S is rolled and the path of P is studied. The initial position of P is (76,0). The initial position of the center of S is (52,0). After S has moved counterclockwise about the origin through an angle t the position of P is x=52cos⁡t+24cos⁡((13/6)t) y=52sin⁡t−24sin⁡((13/6)t) How far does P move before it returns to its initial position? Hint: You may use the formulas for cos( u+v) and sin( w /2). S makes several complete revolutions about the origin before P returns to (76,0).

2496

Evaluate the definite integral ∫5(sqrt(3)/2) 0 sqrt( 25−x^2) dx Answer =

25(4π+3√3)/24

Compute the following limits using l'Hopital's rule if appropriate. Use INF to denote ∞ and -INF to denote −∞. lim x→0. 1−cos(5x)/1−cos(8x) = lim x→1. 9^x−8^x−1/x^2−1 =

25/64 18ln(3)−24ln(2)/2

You and your best friend Janine decide to play a game. You are in a land of make believe where you are a function, f(t), and she is a function, g(t). The two of you move together throughout this land with you (that is, f(t) ) controlling your East/West movement and Janine (that is, g(t) ) controlling your North/South movement. If your identity, f(t), is given by f(t)=(t^2+54)^(3/2)/3 and Janine's identity, g(t), is given by g(t)=27t then how many units of distance do the two of you cover between the Most Holy Point o' Beginnings, t=0, and The Buck Stops Here, t=42? We travel ___ units of distance.

25830

Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If it diverges to infinity, state your answer as "infinity". If it diverges to negative infinity, state your answer as "-infinity". If it diverges without being infinity or negative infinity, state your answer as "divergent". ∫28 1 2/sqrt^3(x−1)dx =

Suppose that f(x)=3^(x^(3)+3x^(2)+5x).Find f′(1). f′(1) =

275562ln(3)

Let L be the circle in the x-y plane with the center of the origin and radius 95. Let S be a moveable circle with a radius of 70. S is rolled along the inside of L without slipping while L remains fixed. A point P is marked on S before S is rolled and the path of P is studied. The initial position of P is (95,0). The initial position of the center of S is (25,0). After S has moved counterclockwise about the origin through an angle t the position of P is x=25cos⁡t+70cos⁡((5/14)t) y=25sin⁡t−70sin⁡((5/14)t) How far does P move before it returns to its initial position? Hint: You may use the formulas for cos( u+v) and sin( w /2). S makes several complete revolutions about the origin before P returns to (95,0).

2800

If f(x)=7arctan⁡(2sin⁡(2x)), find f′(x).

28cos(2x)/4sin^2(2x)+1

Evaluate the limit using L'Hospital's rule lim x→0 9^x−7^x/x

2ln(3)−ln(7)

Let y=x^(log7⁡(x)) Then dy/dx =

2log(x)x^(log(x)/log(7))/xlog(7)

If f is a quadratic function such that f(0)=1 and ∫f(x)/(x^2(x+1)^3)dx is a rational function, find the value of

Find the nth term of the geometric sequence whose initial term is 3 and common ration is 4. (Your answer must be a function of n.)

3 * (4)^(n-1)

Evaluate the definite integral. ∫1 0. 3x^2e^(−4x) dx Answer =

3(2e^4−26)/64e^4

Find the length of the given curve. x=3y^(4/3)−(3y^(2/3))/32, −343≤y≤512 L =

3*8^4+3*8^2/32 + (3*7^4+3*7^2/32)

You and your best friend Janine decide to play a game. You are in a land of make-believe where you are a function, f(t), and she is a function, g(t). The two of you move together throughout this land with you (that is, f(t) ) controlling your East/West movement and Janine (that is, g(t) ) controlling your North/South movement. If your identity, f(t), is given by f(t)=((t^2+14)^(3/2))/3 and Janine's identity, g(t), is given by g(t)=7t then how many units of distance do the two of you cover between the Most Holy Point o' Beginnings, t=0, and The Buck Stops Here, t=21? We travel ___ units of distance.

3234

Use integration by parts to evaluate the integral. ∫8 1 sqrt(t)ln⁡tdt

32√2ln(2)+(−64√2+4/9)

Let L be the circle in the x-y plane with center the origin and radius 95. Let S be a moveable circle with radius 60 . S is rolled along the inside of L without slipping while L remains fixed. A point P is marked on S before S is rolled and the path of P is studied. The initial position of P is (95,0). The initial position of the center of S is (35,0) . After S has moved counterclockwise about the origin through an angle t the position of P is x=35cos⁡t+60cos⁡((7/12)t) y=35sin⁡t−60sin⁡((7/12)t) How far does P move before it returns to its initial position? Hint: You may use the formulas for cos( u+v) and sin( w /2). S makes several complete revolutions about the origin before P returns to (95,0).

3360

Find the length of parametrized curve given by x(t)=9t^2+18t, y(t)=−t^3−3t^2+24t where t goes from 0 to 1. Answer:

If a sequence c1,c2,c3,... has limit K then the sequence e^c1, e^c2, e^c3,... has limit e^K. Use this fact together with l'Hopital's rule to compute the limit of the sequence given by bn=(1+(5.9/n))^n.

365.037

Find the length of the given curve. x=3y^(4/3)−(3y^(2/3)/32), −125≤y≤125 L =

3754.69

39.27

Consider the function f(x)=x/1−3x. Give the Taylor series for f(x) for values of x near 0. ∞ f(x)=∑ ____ n=0 Give the open interval of convergence for this Taylor series. (Note: Give your answer in interval notation.) Give the series expansion of f(x) for very large values of x (i.e., in powers of 1/x). ∞ f(x)=∑ ____ n=0 Give the domain of convergence of this series. (Note: Give your answer in interval notation.)

3^n*x^(n+1) (-0.333333,0.333333) -1/[3^(n+1)]*(1/x)^n (-∞,-0.333333) U (0.333333,∞)

If f(x)=3sin⁡(x)+3x^x, find f′(2). f′(2)=

3cos(2)+12(ln(2)+1)

Evaluate the integral ∫3x^2−6x−3/(x−1)^2(x^2+1)dx Note: Use an upper-case "C" for the constant of integration.

3ln|x−1|+(3/x−1)−(3/2)ln|x^2+1|+3arctan(x)+C

For the following alternating series, ∞ ∑ an=0.6−((0.6)^3/3!)+((0.6)^5/5!)−((0.6)^7/7!)+... n=1 how many terms do you have to compute in order for your approximation (your partial sum) to be within 0.0000001 from the convergent value of that series?

The region bounded by x^2+(y−1)^2=1 is rotated about the y-axis. Find the volume of the resulting solid by any method. Volume =

4.189

Split into partial fractions: 16/(4−x^2)=

4/(2+x) + 4/(2−x)

Find the length of the given curve. x=3y^(4/3)−(3y^(2/3)/32), −27≤y≤216 L =

4135.22

Find the length of the curve defined by y=2x^(3/2)+3. from x=4 to x=10.

47.63

If x=8cos^3⁡θ and y=8sin^3⁡θ, find the total length of the curve swept out by the point (x,y) as θ ranges from 0 to 2π. Answer:

For the following indefinite integral, find the full power series centered at x=0 and then give the first 5 nonzero terms of the power series. f(x)=∫(e^(4x)−1)/7x dx ∞ f(x)=C+∑ ____ n=1 f(x)=C+ __ + __ +__ +___ +

4^n*x^n/[n*(n!)*7] C + 4x/7 + (4^2x^2)/2(2!)7 + (4^3x^3)/3(3!)7 + (4^4x^4)/4(4!)7 + (4^5x^5)/5(5!)7

Let f(x)=x^4tan^−1⁡(7x) f′(x)=

4x^3arctan(7x)+(7x^4/49x^2+1)

Use integration by parts to evaluate the indefinite integral. ∫4arctan⁡(2y)dy Answer =

4ytan^−1(2y)−ln|4y^2+1| +C

Let Tk(x): be the taylor polynomial of degree k of the function f(x)=sin⁡(x) at a=0. Suppose you approximate f(x) by Tk(x), and if |x|≤1, how many terms do you need (that is, what is k) for you to have your error to be less than 1/5040 ? (Hint: use the alternating series approximation.)

Compute the following limits using l'Hospital's rule if appropriate. Use INF to denote ∞ and MINF to denote −∞. lim x→∞. ln⁡(x^5−8)/ln⁡(x)cos⁡(1/x) = lim x→∞. e^(8x)/e^(9x)−e^(−9x) =

5 0

Evaluate the indefinite integral. ∫5sin^−1⁡(3x)dx Answer =

5(xarcsin(3x)+1/3√(1−9x^2))

Use integration by parts to evaluate the integral. ∫5xsin⁡(x)dx=

5(−xcos(x)+sin(x)) + C

5.74349

Find the 4th term of the geometric sequence 6,27,121.5,... Answer:

546.75

Consider the parametric curve given by the equations x(t)=t^2+18t−45y (t)=t^2+18t+6 How many units of distance are covered by the point P(t)=(x(t),y(t)) between t=0 and t=2 ? Answer:

56.57

A cable hangs between two poles of equal height and 31 feet apart. At a point on the ground directly under the cable and x feet from the point on the ground halfway between the poles the height of the cable in feet is h(x)=10+(0.2)(x^1.5). The cable weighs 14.8 pounds per linear foot. Find the weight of the cable.

593.414

Evaluate the limit using L'Hospital's rule if necessary lim x→∞. (7x)^(ln⁡5+1/ln⁡(12x)+1)

Let f(x)=x/(x−5). Find f^−1(x). f^−1(x)= .

5x/(x-1)

Find the length of the curve defined by y=3ln⁡((x/3)^2−1) from x=4 to x=7.

6.09

Compute the following limits using l'H\^opital's rule if appropriate. Use INF to denote ∞ and MINF to denote −∞. lim x→0. 1−cos⁡(8x)/1−cos⁡(9x) = lim x→1. 10^x−9^x−1/x^2−1 =

64/81 10ln(10)−18ln(3)/2

Let f(x)=(6/x)−3e^x. Enter an antiderivative of f(x)

6ln(|x|)−3e^x

Evaluate the limit using L'Hospital's rule if necessary lim x→∞. (1+2/x)^x/12

6√e

How many terms of the series do we need to add in order to find the sum to the indicated accuracy? (Your answer must be the smallest possible integer.) ∞ ∑ (−1)^(n−1)(2/n^4), |error|<0.0005 n=1 Term: n=

Evaluate the integral ∫−7x+21/(x^2+2x+4)^2dx Note: Use an upper-case "C" for the constant of integration.

7/2(x^2+2x+4)+(14/3√3)(arctan((1/√3)(x+1))+(1/2)sin(2arctan((1/√3)(x+1))))+C

Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If not, state your answer as divergent. ∫1 −∞ 7/(2x−3)^2dx

7/2

A cable hangs between two poles of equal height and 35 feet apart. At a point on the ground directly under the cable and x feet from the point on the ground halfway between the poles the height of the cable in feet is h(x)=10+(0.3)(x^1.5). The cable weighs 12.5 pounds per linear foot. Find the weight of the cable.

714.86

Find the length of the given curve. x=3y^(4/3)−(3y^(2/3)/32), −216≤y≤216 L =

7782.75

Find the length of the given curve. x=3y^(4/3)−(3y^(2/3))/32, −64≤y≤343 L =

7977.094

Evaluate the following limit: lim x→∞ 8xe^(1/x)−8x Enter -inf if your answer is −∞, enter inf if your answer is ∞, and enter DNE if the limit does not exist. Limit =

Find the degree 3 Taylor polynomial T3(x) of function f(x)=(−3x+10)^3/2. at a=2. T3(x)=

8 + -9 * (x-2) + 3.375/2 * (x-2)^2 + 1.265625/6 * (x-2)^3

Find ∫16/(4−x^2)dx=

8((1/2)ln|x2+1|−(1/2)ln|x2−1|)+C

If f(x)=8sin⁡(4x)arcsin⁡(x), find f′(x). f′(x) =

8(4cos(4x)arcsin(x)+(sin(4x)/√1−x^2))

Evaluate the indefinite integral. ∫16cos^2⁡(6x)dx

8(x+(1/12)sin(12x)) +C

Find the arc length of the curve y=1/8(x^2−8ln⁡(x)) from x=4 to x=5. Length =

8*ln(5/4)/8+(5^2-4^2)/8

Find the arc length of the curve y=1/8(−x^2+8ln⁡(x)) from x=3 to x=8. Length =

8*ln(8/3)/8+(8^2-3^2)/8

Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If not, state your answer as "divergent". ∫∞ 4 4/(x+7)^(3/2)dx =

8/11^(1/2)

Evaluate the definite integral. ∫π/2 0 sin^5⁡xcos^18⁡x dx Answer:

8/9177

A cable hangs between two poles of equal height and 33 feet apart. At a point on the ground directly under the cable and x feet from the point on the ground halfway between the poles the height of the cable in feet is h(x)=10+(0.5)(x^1.5). The cable weighs 11.2 pounds per linear foot. Find the weight of the cable.

848.64

Find the length of the curve defined by y=4x^(3/2)+9 from x=1 to x=8.

86.81

Evaluate the integral ∫(4x^2−4x+24)/(x^3+3x)dx Note: Use an upper-case "C" for the constant of integration.

8ln|x|−2ln|x^2+3|−(4/√3)arctan(x/√3)+C

Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If not, state your answer as "divergent". ∫∞ 0 9e^(−x)dx =

You and your best friend Janine decide to play a game. You are in a land of make-believe where you are a function, f(t), and she is a function, g(t). The two of you move together throughout this land with you (that is, f(t) ) controlling your East/West movement and Janine (that is, g(t) ) controlling your North/South movement. If your identity, f(t), is given by f(t)=((t^2+6)^3/2)/3 and Janine's identity, g(t), is given by g(t)=3t then how many units of distance do the two of you cover between the Most Holy Point o' Beginnings, t=0, and The Buck Stops Here, t=14? We travel ___ units of distance.

956.67

If x=16cos^3⁡θ and y=16sin^3⁡θ, find the total length of the curve swept out by the point (x,y) as θ ranges from 0 to 2π. Answer:

9^(8/9)

Use integration by parts to evaluate the integral. ∫27x^2cos⁡(3x)dx

9x^2sin(3x)−2(−3xcos(3x)+sin(3x)) + C

For the following function, find the full power series centered at x=0 and then give the first 5 nonzero terms of the power series and the open interval of convergence. f(x)=9/7−x ∞ f(x)=∑ ___ n=0 f(x)=___+ ___+ ___+ ____+ ____+⋯ The open interval of convergence is: (Give your answer in interval notation.)

9x^n/7^(n+1) f(x)= 9x^0/7^(0+1) + 9x^1/7^(1+1) + 9x^2/7^(2+1) + 9x^3/7^(3+1) + 9x^4/7^(4+1) (-7,7)

Rewrite the expression 2log⁡x−4log⁡(x^2+1)+5log⁡(x−1) as a single logarithm log⁡A. Then the function A=

A = (x^2(x-1)^5)/(x^2+1)^4

Write out the form of the partial fraction decomposition of the function: Q=∫10 2 5x/(x^2+4x+4)dx Determine the numerical values of the coefficients, A and B, where B≤A and 5x/(x^2+4x+4)=(A/denominator)+(B/denominator) A= B=

A = 5 B = -10

f(x)=(1/3)x+7, −5≤x≤3 The domain of f^−1 is the interval [A,B] where A= and B=

A = 5.33333 B = 8

An amount of $3000 is invested at an interest rate of 9% per year, compounded quarterly. Find the value A(t) of the investment after t years. A(t)= ____dollars Note: remember to express your function in the variable t. Make a graph of A(t). Which graph most closely resembles the graph of A(t)?Note: you can click on the graphs to get a larger view. Does each graph go to 40 on the x-axis and 24000 on the y-axis.A. B. C. D. ? Use your graph of A(t) to determine when this investment will amount to $20,000. Give your answer in years, correct to the nearest hundredth. You will have to zoom on your graph to determine the answer. ____years

A(t)= 3000*1.0225^4t D 21.3154

The form of the partial fraction decomposition of a rational function is given below. −(3x^2+10x+45)/(x−5)(x^2+9)=(A/x−5)+(Bx+C)/(x^2+9) A= B= C= Now evaluate the indefinite integral. ∫−(3x^2+10x+45)/(x−5)(x^2+9)dx=

A= -5 B= 2 C= 0 −5ln(|x−5|)+ln(|x^2+9|)+C

The inverse of the function f is (A, B or C):

The graph of the function f(x)=−ex can be obtained from the graph of g(x)=ex by one of the following actions: (a) reflecting the graph of g(x) in the y-axis; (b) reflecting the graph of g(x) in the x-axis; Your answer is (input a or b) = The range of the function f(x) is f(x)<A, find A. The value of A = is the domain of the function f(x) still (−∞,∞)? Your answer is (input Yes or No) =

B The value of A = 0 yes

Let f(x)=e−x−8 and g(x)=ex. Consider the following actions: (a) reflecting the graph of g(x) in the x-axis; (b) reflecting the graph of g(x) in the y-axis; (c) shifting the resulting graph to the right 8 units; (d) shifting the resulting graph to the left 8 units; (e) shifting the resulting graph upward 8 units; (f) shifting the resulting graph downward 8 units; The graph of f(x) can be obtained from the graph of g(x) by first doing ___ followed by ___ What is What is the domain of the function f(x)? Give your answer in interval notation. What is the range of the function f(x)? Give your answer in interval notation.

B followed by F Domain = (-inf,inf) Range = (-8,inf)

Determine whether the following series converges or diverges. ∞ ∑ (−1)^(n+1)/3n^6+3 n=1 Input C for convergence and D for divergence:

Determine whether the following series converges or diverges. ∞ ∑ (−1)^(n−1)(sqrt(n)/n+7) n=1 Input C for convergence and D for divergence:

Determine whether the following series converges or diverges. ∞ ∑ (−1)^(n−1)/n n=1 Input C for convergence and D for divergence:

Determine whether the following series converges or diverges. ∞ ∑ (−1)^nsin⁡(7/n) n=1 Input C for convergence and D for divergence:

Determine whether the following series converges or diverges. ∞ ∑ cos⁡(nπ)/n^(2/6) n=1 Input C for convergence and D for divergence:

What is the correct form of the partial fraction decomposition for the following integral? ∫x^2+1/(x−3)^3(x^2+9x+45)dx A. ∫((A/x−3)+(Bx+C/(x−3)^2 )+(Dx+E/(x−3)^3)+((Fx+G)/(x^2+9x+45))dx B. There is no partial fraction decomposition because the denominator does not factor. C. There is no partial fraction decomposition yet because there is cancellation. D. ∫((A/x−3)+(B/(x−3)^2)+(C/(x−3)^3)+((Dx+E)/(x^2+9x+45))dx E. ∫((A/(x−3)^3)+(B/x−9)+(C/(x−9)^2)+((Dx+E)/(x^2+1))dx F. ∫((A/(x−3)^3)+(B/x−9)+(C/x−45))dx G. ∫((A/(x−3)^3)+((Bx+C)/(x^2+9x+45))dx H. There is no partial fraction decomposition yet because long division must be done first.

The improper integral ∫∞ −∞ xdx is, A. convergent since the area to the left of x=0 cancels with the area to the right of x=0. B. divergent by comparison to ∫∞ −∞ xe^(−x)dx. C. convergent since it equals lim a→−∞∫0 a xdx+lim b→∞∫b 0 xdx= −∞+∞=0. D. divergent since ∫0 −∞ xdx is convergent and ∫∞ 0 xdx is divergent. E. divergent since both integrals ∫0 −∞. xdx=−∞ and ∫∞ 0. xdx=+∞ are divergent. F. divergent by comparison to ∫∞ −∞ xdx. G. convergent since it equals lim t→∞∫t −t xdx= lim t→∞(t^2/2−((−t)^2)/2)=0.

Sketch the solid obtained by rotating the region underneath the graph of the function f(x)=x^−4 over the interval [−4,−1] about the axis x=4 and calculate its volume using the Shell Method. V =

Each shell has radius 4−x and height x^−4, so the volume of the solid is 2π∫−1 -4. (4−x)(x^−4)dx 2π (1/2x^−2−4/3x^−3)/−1 -4 =57π/16

The half-life of Radium-226 is 1590 years. If a sample contains 500 mg, how many mg will remain after 1000 years? Hint: Recall P=P0e^kt where P0 is the initial amount of such a substance, P is the amount at a given time t and k is the exponential rate of decay. What would it mean if P=.5P0?

Find k by setting .5P0=P0e^1590k, since 1590 is the half-life. Solve this for k and use this and the given 500 = P0 to find that P = 323.327737322167.

Human hair from a grave in Africa proved to have only 82% of the carbon 14 of living tissue. When was the body buried? The half life of carbon 14 is 5730 years. The body was buried about ___ years ago. Hint: The half-life of Carbon-14 is 5730 years. Use this and the the information about 82 % to help you find r. Then find t.

Find r by setting .5P0=P0e^r5730, since 5730 is the half-life. Solve this for r and use this and the given 82% to set up .0.82P0=P0e^rt and solve for t to find that t = 1641.

Evaluate the integral ∫dx/sqrt(20x+10x^2) by completing the square and using trigonometric substitution.

First complete the square: 10x^2+20x=10(x^2+2x+1−1)=10(x+1)^2−10. Let u=x+1. Then, du=dx, and sqrt(20x+10x^2)= sqrt(10)sqrt(u^2−1)I= ∫dx/sqrt(20x+10x^2)= (1/sqrt(10))∫du/sqrt(u^2−1). Now let u=1sec⁡t. Then du=1sec⁡t tan ⁡t dt, and u^2−1=1sec^2⁡t−1=1(sec^2⁡t−1)=1tan^2⁡t. Thus, I=(1/sqrt(10))∫du/sqrt(u^2−1)= (1/sqrt(10))∫(1sec⁡ttan⁡tdt)/1tan⁡t= (1/sqrt(10))∫sec⁡tdt= (1/sqrt(10))ln⁡|sec⁡t+tan⁡t|+C. Since u=1sec⁡t, we construct a right triangle with sec⁡t=u1. Here a=1 From that triangle, the Pythagorean theorem, and the definition of secant with respect to right triangles, we see that tan⁡t=sqrt(u^2−1)/1, so we have I=sqrt(1/10)ln⁡|(u/1)+sqrt(u^2−1)/1|+C= sqrt(1/10)ln⁡|u+sqrt(u^2−1)|+C= sqrt(1/10)ln⁡|(x+1)+sqrt(2x+x^2)|+C because sqrt(20x+10x^2) = sqrt(10)sqrt(u^2−1).

Find the local quadratic approximation of f at x=x0, and use that approximation to find the local linear approximation of f at x0. Use a graphing utility to graph f and the two approximations on the same screen. f(x)=sin(x), x0=π/2 Enter Approximation Formulas below. Local Quadratic Approx = Local Linear Approx =

Let f(x)=sin(x), f′(x)=cos(x), f″(x)=−sin(x). f(x0)=1, f′(x0)=0,f″(x0)=−1 Local Quadratic Approximation: f(x)=sin(x)≈f(x0)+f′(x0)(x−x0)+(f″(x0)/2)(x−x0)^2= 1−(1/2)(x−(π/2))^2 Local Linear Approximation: f(x)=sin(x)≈f(x0)+f′(x0)(x−x0)= 1

Use integration by parts to evaluate the integral. ∫7xln⁡(3x)dx

Let u = ln⁡(3x) and dv = 7xdx. Then du = (1/3x)⋅3dx = (1/x)dx and v = 3.5x^2. ∫7xln⁡(3x)dx= uv−∫vdu = ln⁡(3x)3.5x^2−∫3.5x^2(1/x)dx = 3.5x^2ln⁡(3x)−∫3.5xdx = 3.5x^2ln⁡(3x)−1.75x^2+C

Use the substitution x=6sec⁡t to evaluate the integral ∫dx/xsqrt(x^2−36). Note: Use C for an arbitrary constant.

Let x=6sec⁡t. Then dx=6sec⁡ t tan ⁡t dt, and sqrt(x^2−9)= sqrt(36sec^2⁡t−36)= 6sqrt(sec^2t−1)= 6sqrt(tan^2⁡t)= 6tan⁡t Thus, I=∫dx/xsqrt(x^2−36)= ∫(6sec⁡ttan⁡tdt)/(6sec⁡t)(6tan⁡t)= (1/6)∫dt= (1/6)t+C. Since x=6sec⁡t, t=sec−1⁡(x6), and ∫dx/xsqrt(x^2−36)= (1/6)sec^−1⁡(x/6)+C.

Use the substitution x=6sin⁡t to evaluate the integral ∫sqrt(36−x^2)dx. Note: Use C for an arbitrary constant.

Let x=6sin⁡t. Then dx=6cos⁡tdt, and I=∫sqrt(36−x^2)dx= ∫sqrt(36−36sin^2⁡t)(6cos⁡tdt)= 36∫sqrt(1−sin^2⁡t)cos⁡tdt= 36∫sqrt(cos^2⁡t)cos⁡tdt= 36∫cos^2⁡tdt= 36[(1/2)t+(1/2)sin⁡tcos⁡t]+C= (36/2)t+(36/2)sin⁡tcos⁡t+C Since x=6sin⁡t, we construct a right triangle with sin⁡t=x/6. Here a=6 From that triangle, the Pythagorean theorem, and the definition of cosine with respect to right triangles, we see that cos⁡t=1636−x2, so we have I=(36/2)t+(36/2)sin⁡tcos⁡t+C= (36/2)sin^−1⁡(x/6)+(36/2)(x/6)(sqrt(36−x^2)/6)+C= 18sin^−1(x/6)+xsqrt(36−x^2)2+C.

Newton's Law of Cooling states that the rate at which an object cools is proportional to the difference in temperature between the object and the surrounding medium. Thus, if an object is taken from an oven at 310∘F and left to cool in a room at 78∘F, its temperature T after t hours will satisfy the differential equation dT/dt=k(T−78). If the temperature fell to 202∘F in 0.7 hour(s), what will it be after 3 hour(s)? After 3 hour(s), the temperature will be ____ degrees F. Hint: Newton's Law of Cooling is discussed in the book on pages 240--241.

Let y=T−78 to turn the equation into y′=ky. As with the other problems in this section, this leads to the solution y=Ce^kt. When t=0, T=310, thus we can solve for C. When t=0.7,T=202, thus we can solve for k. Now the formula is given in terms of t. Plug in t=3 to see y=, and therefore T=93.8303150259589∘.

Find the local quadratic approximation of f at x=x0, and use that approximation to find the local linear approximation of f at x0. Use a graphing utility to graph f and the two approximations on the same screen. f(x)=e^−7x, x0=0 Enter Approximation Formulas below. Local Quadratic Approx = Local Linear Approx =

Local Quadratic Approx = 1-7*x+49/2*x^2 Local Linear Approx = 1-7*x

Evaluate the indefinite integral. ∫ 13sqrt(x^2−16)/x dx =

Make the substitution x=4sec⁡(θ). Then dx=4sec⁡(θ)tan⁡(θ)dθ, and sqrt(x^2−16)=sqrt(16sec^2⁡(θ)−16)=sqrt(16tan^2⁡(θ))=4tan⁡(θ). After substitution, the integral becomes ∫13sqrt(x^2−16)/x dx= 13∫(4tan⁡(θ)/4sec⁡(θ))4sec⁡(θ)tan⁡(θ) dθ= 4⋅13∫tan^2⁡(θ) dθ= 4⋅13∫(sec^2⁡(θ)−1) dθ= 4⋅13(tan⁡(θ)−θ)+C for some constant C. From the picture of a right triangle in which x=4sec⁡(θ), or equivalently 4=xcos⁡(θ), we obtain the substitutions 4⋅13(tan⁡(θ)−θ)+C= 4⋅13((sqrt(x^2−16)/4)−arccos⁡(4/x))+C= 13sqrt(x^2−16)−4⋅13arccos⁡(4/x)+C.

Let f(x)=x^7x. Use logarithmic differentiation to determine the derivative. f′(x)=

Method 1: x^7x = e^(7xln⁡x), so d/dx x^7x = e^(7xln⁡x)(7+7ln⁡x) = x^7x(7+7ln⁡x). Method 2: Let y = x^7x. Then, ln⁡y = 7xln⁡x. By logarithmic differentiation y′/y = 7x⋅(1/x)+7ln⁡x, so y′ = y(7+7ln⁡x) = x^7x(+7ln⁡x).

Approximate the value of the series to within an error of at most 10^−3. ∞ ∑ (−1)^(n+1)/(n+9)(n+6) n=1 According to Equation (2): |SN−S|≤aN+1 what is the smallest value of N that approximates S to within an error of at most 10^−3? N= S≈

N= 24 S≈ 0.00751

Part 1: Use differentiation and/or integration to express the following function as a power series (centered at x=0). f(x)=1/(3+x)^2 ∞ f(x)=∑ n=0 Part 2: Use your answer above (and more differentiation/integration) to now express the following function as a power series (centered at x=0). g(x)=1/(3+x)^3 ∞ g(x)=∑ n=0 Part 3: Use your answers above to now express the function as a power series (centered at x=0). h(x)=x^2/(3+x)^3 ∞ h(x)=∑ n=0

Part 1: ((-1)^n(n+1)x^n)/(3^(n+2)) Part 2: ((-1)^n(n+1)(n+2)X^n)/2(3^(n+3)) Part 3: ((-1)^n(n+1)(n+2)x^(n+2))/2(3^(n+3))

PART 1: Use the partial fractions method to express the function as a power series (centered at x=0) and give the open interval of convergence. f(x)=9/(x^2+1x−20) ∞ f(x)=∑ n=0 The open interval of convergence is: Give your answer in interval notation. PART 2: Now use the method of completing the square to express the function as a power series and give the natural center and open interval of convergence. f(x)=9/(x^2+1x−20) ∞ f(x)=∑ n=0 The center is: The open interval of convergence is: Give your answer in interval notation.

Part 1: [-0.2^(n+1)-0.25^(n+1)]x^n (-4,4) Part 2: (-9/20.25^(n+1))(x+0.5)^2n -0.5 (-5,4)

Let f(x)=9^(x)log3⁡(x) f′(x)=

Recall that (ax)′=ln⁡(a)a^x and (loga⁡(x))′=1/ln⁡(a)x. We will also need to use the identity loga⁡(x)=ln⁡(x)/ln⁡(b) because we can't enter loga⁡(x) in WeBWorK. Using the product rule we have f′(x)=ln⁡(9)9^xlog3⁡(x)+9^x(1/ln⁡(3)x) = (ln⁡(9)9^xln⁡(x)/ln⁡(3))+(9^x/ln⁡(3)x).

Consider the error in using the approximation e−x≈1−x on the interval [−1,1]. (a) Reasoning informally, on what interval is this approximation an overestimate? An underestimate? (For each, give your answer as an interval or list of intervals, e.g., to specify the intervals −0.25≤x<0.5 and 0.75<x≤1, enter [-0.25, 0.75), (0.75,1] Enter none if there are no such intervals.) (b) Use the Error Bound for Taylor Polynomials to find a good smallest bound for the error in approximating e−x with 1−x on this interval: error bound = Now, consider the error in using the approximation e−x≈1−x+(x^2)/2!−(x^3)/3! on the same interval. (c) Reasoning informally, on what interval is this approximation an overestimate? An underestimate? (For each, give your answer as an interval or list of intervals, e.g., to specify the intervals −0.25≤x<0.5 and 0.75<x≤1, enter [-0.25, 0.75), (0.75,1] Enter none if there are no such intervals.) (d) Use the Error Bound for Taylor Polynomials to find a good smallest bound for the error in approximating e−x with 1−x+(x^2)/2!−(x^3)/3! on this interval: error bound =

SOLUTION (a) 1−x is the first degree approximation of f(x)=e^−x. P1(x)=1−x is an overestimate for no values of x and an underestimate for [−1,0) and (0,1]. (This can be seen easily from a graph.) (b) Using the first degree error bound, if |f^(2)(x)|≤M for −1≤x≤1, then |E1|≤(M⋅|x|^2/2!)≤M/2. For what value of M is |f(2)(x)|≤M for −1≤x≤1? Well, |f^(2)(x)|=|e^−x|≤e^0.1. So |E1|≤e^0.1/2. (c) As before, we note that 1−x+(x^2)/2!−(x^3)/3! is the 3th degree approximation of f(x)=e^−x. This is an overestimate for no values of x and an underestimate for [−1,0) and (0,1]. (This also can be seen easily from a graph.) (d) Again, similar to (b), our error bound is |E3|≤M⋅|x|^4/(4)!≤M/24, where |f(^4)(x)|≤M for −1≤x≤1. Then a good value for M is M=e^0.1, so |E3|≤e^0.1/24.

Find the Maclaurin polynomials of orders n=0,1,2,3, and 4, and then find the nth Maclaurin polynomials for the function in sigma notation. Enter the Maclaurin polynomials below for 9/1+x. p0(x)= p1(x)= p2(x)= p3(x)= p4(x)= n pn(x)= ∑ m=0

SOLUTION Let f(x)=9/1+x, then f′(x)=−9/(1+x)^2, f″(x)=18/(1+x)^3, f‴(x)=−54/(1+x)^4, f^(4)(x)=216/(1+x)^5 Then f(0)=9, f′(0)=−9, (f″(0)/2)=9, (f‴(0)/6)=−9, (f^(4)(0)/24)=9 Then we have, p0(x)=f(0) = 9 p1(x)=f(0)+f′(0)x = 9−9x p2(x)=f(0)+f′(0)x+(f″(0)/2)x^2 = 9−9x+9x^2 p3(x)=f(0)+f′(0)x+(f″(0)/2)x^2+(f‴(0)6)/x^3 = 9−9x+9x^2−9x^3 p4(x)=f(0)+f′(0)x+(f″(0)2)/x^2+(f‴(0)6)/x^3+(f^(4)(0)/24)x^4 = 9−9x+9x^2−9x^3+9x^4 n pn(x)=9∑ (−1)^mx^m m=0

Find the Taylor polynomial of degree 3 around the point x=1 of f(x)=sqrt^3(4+x). P3(x)=

SOLUTION Let f(x)=sqrt^3(4+x)=(4+x)^1/3. Then f′(x)=0.333333(1/(4+x)^0.666667), f''(x)=−0.333333(0.666667(1/(4+x))^0.333333)/((4+x)^0.666667)^2, and f'''(x)=0.333333(0.666667(0.333333(1/(4+x)^0.666667))/((4+x)^0.333333)^2)((4+x)^0.666667)^2+0.666667(1/(4+x)^0.333333)⋅2(4+x)^0.666667⋅0.666667(1/(4+x)^0.333333))/(((4+x)0.666667)^2)^2. The Taylor polynomial of degree three about x=1 is thus P3(x)=(4+1)^1/3+(1/3(4+1)^2/3)(x−1)+(1/2!)(−2/(9(4+1)^5/3))(x−1)^2+(1/3!)(−10/(27(4+1)^8/3))(x−1)^3.

Given the equation: xy=10, set up an integral to find the length of path from x=a to x=b and enter the integrand below. (i.e. if your integral is L=∫b a c^2x^2/h dx enter c^2x^2/h as your answer.) L=∫b a ___ dx

SOLUTION xy=10⇒y=10/x⇒dy/dx=−10/x^2 Thus L=∫b a sqrt(1+(dy/dx)^2)dx= ∫b a sqrt(1+(−10/x^2^)2)dx= ∫b a. sqrt(1+(100/x^4))dx

Consider the convergent alternating series ∞ ∑ (−1)^n/n!=L. n=1 Let Sn be the nth partial sum of this series. Compute Sn and Sn+1and use these values to find bounds on the sum of the series. If n=4, then Sn= and Sn+1= and therefore ∞ ___ <∑ (−1)^n/n!< ____ n=1 This interval estimate for the value of the series has error |Sn−L|<.

Sn= -0.625 Sn+1= -0.6333 ∞ -0.633333 <∑ (−1)^n/n!< -0.625 n=1 error |Sn−L|< 0.008333

Compute the surface area of revolution of y=4x+3 about the x-axis over the interval [4,6].

Solution: S=2π∫6 4 (4x+3)sqrt(17)dx = 92πsqrt(17)

Compute the surface area of revolution of y=(4−x^(2/3))^(3/2) about the x-axis over the interval [0,2].

Solution: y′=(3/2)sqrt(4−x^(2/3))(−(2/3)x^(−1/3)) 1+(y′)2= 4/x^(2/3) S=2π∫2 0 (4−x^(2/3))^(3/2)(2/x^(1/3))dx We use the substitution u=4−x^(2/3), du=−2/3x^(1/3) to get: S=2π(−6(4−x^(2/3))^(5/2))/5 |2 0. =173.107

Approximate the value of the series to within an error of at most 10^−5. ∞ ∑ (−1)^(n+1)/n^7 n=1 According to Equation (2):|SN−S|≤aN+1 what is the smallest value of N that approximates S to within an error of at most 10−5? N= S≈

Solution: Let ∞ S=∑ (−1)^(n+1)/n^7, n=1 so that an=1/n^7. By Equation (2), |SN−S|≤aN+1=1/(N+1)^7. To make the error less than 10^−5, we must choose N so that 1/(N+1)^7 < 10^−5 or N > 10^(5/7)−1 ≈ 4.18. The smallest value that satisfies the required inequality is then N=5. Thus, S ≈ S5 = 1−(1/2^7)+(1/3^7)−(1/4^7)+(1/5^7) = 0.992597

Calculate the Taylor polynomials T2(x) and T3(x) centered at x=π6 for f(x)=cos(x). T2(x) must be of the form A+B(x−(π/6))+C(x−(π/6))^2 where A equals: B equals: C equals: T3(x) must be of the form D+E(x−(π/6))+F(x−(π/6))^2+G(x−(π/6))^3 where D equals: E equals: F equals: G equals:

Solution: Recall the general formula for the Taylor polynomial centered at x = a: T(x)=f(a)+(f′(a)/1!)(x−a)+(f″(a)/2!)(x−a)^2+...+(fn(a)/n!)(x−a)^n So, in this case, f(x)=cos(x) and T2(x)=A+B(x−(π/6))+C(x−(π/6))^2 where A=f(π6) = sqrt(3)/2 = 0.866025403784439 B=f′(π6) = −1/2 = −0.5 C=f″(π6)2! = −sqrt(3)/4 = −0.433012701892219 Thus, T2(x)=sqrt(3)/2−(1/2)(x−(π/6))−sqrt(3)/4(x−(π/6))^2 or T2(x)=0.866025−0.5(x−(π/6))−0.433013(x−(π/6))^2 Similarly, T3(x)=D+E(x−(π/6))+F(x−(π/6))^2+G(x−(π/6))^3 where D=f(π6) = sqrt(3)/2 = 0.866025403784439 E=f′(π6) = −1/2= −0.5 F=f″(π6)2! = −sqrt(3)/4 = −0.433012701892219 G=f‴(π6)3!= 1/12 = 0.0833333333333333 Thus, T3(x)=sqrt(3)/2−(1/2)(x−(π/6))−sqrt(3)/4(x−(π/6))^2+1/12(x−(π/6))^3 or T3(x)=0.866025−0.5(x−(π/6))−0.433013(x−(π/6))^2+0.0833333(x−(π/6))^3

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.y=1/x^4,y=0,x=4,x=8; about the y-axis. Volume =

The region to be rotated about the y−axis is shown below (not to scale) We use the method of cylindrical shells. The region is rotated around the y−axis, so we have thin vertical cylindrical shells, of thickness dx. The interval needed by x to fill the region bounded by the curves is 4≤x≤8. For the integrand, we think of a cylindrical shell at x with height h=1/x^4, thickness dx, rotated around the line x=0 and so has circumference 2πr=2πx. The volume then is V=∫8 4. 2πx(1/x^4)dx =2π∫8 4(1/x^3)dx =2π[x^−2/−2]8 4 =−1π(1/8^2−1/4^2)(cubic units)

Find the volume of the solid obtained by rotating the region under the graph of the function f(x)=2x−x^2 about the x-axis over the interval [0,2]. V =

The volume V of the solid of revolution is V=π∫2 0(x^2−2x)^2dx = π∫2 0(x^4−2(2)x^3+(2^2)x^2)dx =π(1/5x^5−2/2x^4+2^2/3x^3)|2 0 =π(1/5*(2)^5−2/2*(2)^4+2^2/3*(2)^3) = π/30*(2^5) = 3.35103216382911

Find the limit: lim t→0. (3/t−(3/e^t−1))= (Enter undefined if the limit does not exist.)

To get this limit into a form to which l'Hopital's rule applies, we combine fractions. Then we have lim t→0. 3e^t−3−3t/t(e^t−1)= lim t→0. 3e^t−3/tet^+e^t−1= lim t→0. 3e^t/te^t+2e^t = 3/2.

Find the limit: lim x→0. ((4/x)−(4/sin⁡(x)))=

To get this limit into a form to which l'Hopital's rule applies, we combine fractions. Then we have lim x→0. 4sin⁡(x)−4x/xsin⁡(x)= lim x→0. 4cos⁡(x)−4/sin⁡(x)+xcos⁡(x)= lim x→0. −4sin⁡(x)/2cos⁡(x)−xsin⁡(x) = 0

Find the volume of the solid obtained by revolving the graph of y=6xsqrt(4−x^2) over [0,2] about the y-axis.

Using the method of cylindrical shells, the volume is given by V=2π∫2 0 x(6xsqrt(4−x^2))dx= 12π∫2 0 x^2sqrt(4−x^2)dx. To evaluate this integral, let x=2sin⁡θ. Then dx=2cos⁡θdθ, 4−x^2 = 4(1−sin^2⁡θ) = 4cos^2⁡θ, and I=∫x^2sqrt(4−x^2)dx= ∫(2^2sin^2⁡θ)(2cos⁡θ)(2cos⁡θ)dθ= 2^4∫(1−cos^2⁡θ)cos^2⁡θdθ. (1/2^4)I= ∫cos^2⁡θdθ−∫cos^4⁡θdθ. Now we use the reduction formula for ∫cos^4⁡θdθ: (1/2^4)I=∫cos^2⁡θdθ−[((cos^3θsin⁡θ)/4)+(3/4)∫cos^2⁡θdθ]= −1/4cos^3⁡θsin⁡θ+1/4∫cos^2⁡θdθ= −1/4cos^3⁡θsin⁡θ+1/4[(1/2)θ+(1/2)sin⁡θcos⁡θ]+C= −1/4cos^3⁡θsin⁡θ+1/8θ+1/8sin⁡θcos⁡θ+C. Since sin⁡θ=x/2, we know that cos⁡θ=sqrt(4−x^2)/2. Then we have (1/2^4)I=−1/4(sqrt(4−x^2)/2)^3(x/2)+1/8sin^−1⁡(x/2)+1/8(x/2)(sqrt(4−x^2)/2)+C. Now we can complete the volume calculation: V=12πI|2 0= 12⋅2^4π[−1/4(sqrt(4−x^2)/2)^3(x/2)+1/8sin^−1⁡(x/2)+1/8(x/2)(sqrt(4−x^2)/2)]2 0= 12⋅2^4π(1/8)(π/2)= 6⋅2^4π^2(1/4)= 12π^2.

Find the value of ∫π/3 0 cos⁡(3x)dx. Remember: The angles for sin and cosine are always (well... almost always) in radians!

Using the substitution u=3x (and so du=3dx) we get: ∫π/3 0. cos⁡(3x)dx= (1/3)∫π/3 0. cos⁡(3x)⋅3dx= (1/3)∫π/3 0 cos⁡(u)du= (1/3)(sin⁡(3x))|π/3 0. = (1/3)(sin⁡(3π/3)−sin⁡(0))= (1/3)sin⁡(3π/3)

Use the Shell Method to find the volume of the solid obtained by rotating region above the graph of f(x)=x^2+2 and below y = 6 for 0≤x≤2 about the y-axis. V =

When rotating the region in question about the y-axis, each shell has radius x and height 6-(x^2+2)=4−x^2. The volume of the resulting solid is 2π∫2 0x(4−x^2)dx =2π∫2 0(4x−x^3)dx =2π(4x^2/2−1/4x^4)|2 0 =8π

Suppose that P2(x)=a+bx+cx^2 is the second degree Taylor polynomial for the function f about x=0. What can you say about the signs of a, b, c if f has the graph given below? (For each, enter + if the term is positive, and − if it is negative. Note that because this is essentially multiple choice problem it will not show which parts of your answer are correct or incorrect.) a is b is c is

a = + b = - c = -

(a) If f is one-to-one and f(−11)=5, then f^−1(5)= and (f(−11))^−1= . (b) If g is one-to-one and g(6)=4, then g^−1(4)= and (g(6)^)−1= .

a) then f^−1(5) = -11 and (f(−11))^−1 = 0.2 b) then g^−1(4) = 6 and (g(6)^)−1 = 0.25

Find the area under the curve y=1x^(−2) from x=9 to x=t and evaluate it for t=10 , t=100. Then find the total area under this curve for x≥9. (a) t = (b) t = (c) Total area

a) 0.1111 b) 0.1111 c) 0.1111

Find the area under the curve y=1/(4x^3) from x=1 to x=t and evaluate it for t=10,t=100. Then find the total area under this curve for x≥1. (a) t = (b) t = (c) Total area

a) 0.124 b) 0.125 c) 0.125

Define the functions F(x) and G(x) by F(x)=∫x −x t^7 dt, G(x)=∫x+8 −x+8 t^7 dt Determine whether each of the following improper integrals and limits is divergent or convergent. If it is convergent, evaluate it. If it diverges to infinity, state your answer as "INF" (without quotation marks). If it diverges to negative infinity, state your answer as "MINF". If it diverges without being infinity or negative infinity, state your answer as "DIV". (a) ∫∞ −∞ x^7dx (b) lim x→∞ F(x) (c) lim x→∞ G(x)

a) div b) 0 c) ∞

Find the following limits, using L'H\^opital's rule, if appropriate. Use INF to denote ∞ and MINF to denote −∞ (a) lim x→∞ tan^−1⁡(x/6)/sin^−1⁡(1/x) = (b) lim x→0 xcos^5⁡(πe^(x^10))/ln⁡(1+2x) =

a) ∞ b) −1/2

To find the length of the curve defined by y=2x^5+17x from the point (-2,-98) to the point (2,98), you'd have to compute ∫b a f(x)dx. where a=, b=, and f(x)=.

a= -2 b= 2 f(x)= sqrt(1+((2x^5+17x)')^2) = sqrt(1+(10x^4+17)^2)

To find the length of the curve defined byy=4x5+13xfrom the point (-3,-1011) to the point (2,154), you'd have to compute ∫b a f(x)dx where a= b= and f(x)=

a= -3 b= 2 f(x)= sqrt(1 + (20^(x^(4)) + 13)^2)

To find the length of the curve defined by y=6x^5+16x from the point (0,0) to the point (3,1506), you'd have to compute ∫b a f(x)dx where a= b= and f(x)=

a= 0 b= 3 f(x)=sqrt(1 + (30^(x^(4)) + 16)^2)

Evaluate the integral ∫−3/(x+a)(x+b)dx for the cases where a=b and where a≠b. Note: For the case where a=b, use only a in your answer. Also, use an upper-case "C" for the constant of integration. If a=b: If a≠b:

a=b: -3*(-1/(x+a))+C+c a≠b:-3/(b-a)*ln(abs((x+a)/(x+b)))+C+c

Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If not, state your answer as "divergent." ∫5 −∞ 1/(x^2+1)dx

arctan(5)+(π/2)

A function f(x) is graphed in plane A. It is clearly a 1:1 function, so it must have an inverse. Enter the color ("red", "green", or "blue") of this inverse function which is graphed in-plane B. Use what you know about the graphs of inverse functions rather than algebraic calculations based on what you might guess the function to be. Color of f^−1 graph =

blue

Suppose that f(x) and g(x) are given by the power series f(x)=4+2x+6x^2+2x^3+⋯ and g(x)=4+7x+2x^2+2x^3+⋯. By multiplying power series, find the first few terms of the series for the product h(x)=f(x)⋅g(x)=c0+c1x+c2x^2+c3x^3+⋯. c0 = c1 = c2 = c3 =

c0 = 16 c1 = 36 c2 = 46 c3 = 62

Suppose that f(x) and g(x) are given by the power series f(x)=2+3x+2x^2+2x^3+⋯ and g(x)=8+18x+23x^2+31x^3+⋯. By dividing power series, find the first few terms of the series for the quotient h(x)=g(x)f(x)=c0+c1x+c2x^2+c3x^3+⋯. c0 = c1 = c2 = c3 =

c0 = 4 c1 = 3 c2 = 3 c3 = 4

Use the binomial series to expand the function f(x)=1/(5+x)^3 as a power series. ∞ ∑cnx^n. n=0 Compute the following coefficients: c0= c1= c2= c3= c4=

c0= 0.008 c1= -0.0048 c2= 0.0019 c3= -0.00064 c4= 0.00019

Use the binomial series to expand the function f(x)=(1+x)^1/2 as a power series. ∞ ∑cnx^n. n=0 Compute the following coefficients: c0= c1= c2= c3= c4=

c0= 1 c1= 0.5 c2= -0.125 c3= 0.0625 c4= -0.0391

The fourth degree Taylor polynomial for f(x)=x3 centered at a=1 is T4(x)=c0+c1(x−1)+c2(x−1)^2+c3(x−1)^3+c4(x−1)^4. Find the coefficients of this Taylor polynomial. c0= c1= c2= c3= c4= ___ The function f(x)=x3 equals its fourth degree Taylor polynomial T4(x) centered at a=1. Hint: Graph both of them. If it looks like they are equal, then do the algebra.

c0= 1 c1= 3 c2= 3 c3= 1 c4= 0 True

Find the exact value of the series. ∞ ∑ (−1)^nπ^(2n)/4^(2n)(2n)!= n=0

cos(pi/4)

div

Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If not, state your answer as "divergent". ∫∞ 3 ln(x)/x dx =

divergent

Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If not, state your answer as "divergent". ∫∞ −∞ ((x+4)^2+1)dx

divergent

Find the exact value of the series ∞ ∑ 11^n/13^n(n!)= n=0

e^(11/13)

Find the limit of the sequence whose terms are given by an=(e^(2n)+6n)^(1/n).

e^2

The function f(x) is approximated near x=0 by the second degree Taylor polynomial P2(x)=5−x−8x^2. Give values: f(0)= f′(0)= f″(0)=

f(0)= 5 f′(0)= −1 f″(0)= −16

Let f(x)=3xsin^−1(x). Find f′(x). f′(x)= Find f′(0.4). f′(0.4)=

f′(x) = 3(arcsin(x)+(x/(√1−x^2))) f′(0.4) = 3arcsin(2/5)+(2√21/7)

If f(x)=6arctan⁡(8x), find f′(x). Find f′(1).

f′(x) = 6/(1+(8x)^2) d/dx (8x) = 48/(1+64x^2) f′(1) = 48/(1+64⋅1^2) = 48/65

Let f(x)=ln⁡(x^6) f′(x)= f′(e^4)=

f′(x) = 6/x f′(e^4) =6e^-4

Let f(x)=−2log2⁡(x) f′(x)= f′(3)=

f′(x)= −2/ln(2)x f′(3)= −2/3ln(2)

If f(x)=4x^2arctan⁡(4x^4), find f′(x). f′(x) =

f′(x)=4x^2d/dx(tan^−1(4x^4))+tan^−1(4x^4)d/dx(4x^2) by the product rule = 4x^2(1/1+(4x^4)^2)d/dx(4x^4)+tan^−1(4x^4)(8x)= 4x^2(1/1+(4x^4)^2)(4⋅4x^3)+tan^−1(4x^4)(8x)= 64x^5/(1+16x^8)+8xtan^−1(4x^4)

Let f(x)=7e^(xcos⁡x) f′(x)=

f′(x)=7e^(xcos⁡x)d/dx(xcos⁡x)=7e^(xcos(x))(cos(x)−xsin(x))

Evaluate the integral ∫dx/sqrt(x^2+4x+20) by completing the square and using trigonometric substitution. Answer: Note: Use C for an arbitrary constant.

ln((1/4)|x+2+√(x^2+4x+20)|)+C

Let f(x)=tan^−1⁡(3x) f′(x)=

ln(3)·3^x/9^(x)+1

∫e^3 e^2. 2/(yln⁡(y)) dy=

ln(9/4)

Evaluate the indefinite integral. ∫sec⁡x dx Answer:

ln|tan(x)+sec(x)| + C

Find the Maclaurin polynomials of orders n=0,1,2,3, and 4, and then find the nth Maclaurin polynomials for the function in sigma notation. Enter the Maclaurin polynomials below for 6xsin(x). p0(x)= p1(x)= p2(x)= p3(x)= p4(x)= n pn(x)= ∑ (Note summation starts at m=1). m=1

p0(x)= 0 p1(x)= 0 p2(x)= 6*x^2 p3(x)= 6*x^2 p4(x)= 6*x^2-x^4 pn(x)= 6*(-1)^(m-1)*x^(2*m)/(2*m-1)!

Find the first four Taylor polynomials about x=x0, and use a graphing utility to graph the given function and the Taylor polynomials on the same screen. ln(x+9);x0=−8. p0(x)= p1(x)= p2(x)= p3(x)=

p0(x)= 0 p1(x)= x+8 p2(x)= x+8-1/2*(x+8)^2 p3(x)= x+8-1/2*(x+8)^2+1/3*(x+8)^3

Use calculus to find the volume of a cap of a sphere with height h and radius r. volume =

pi*h^2(r-(h/3))

Let f(x)=x/(x^2+5x+29). A. Find the smallest real number r such that f(x) is decreasing for all x greater than r. r= B. Find the smallest integer s such that f(n) is decreasing for all integers n greater than or equal to s. s=

r = 5.39 s = 5

Note: You can get full credit for this problem by just entering the final answer (to the last question) correctly. The initial questions are meant as hints towards the final answer and also allow you the opportunity to get partial credit. Consider the definite integral ∫1/4 0 xsin^−1⁡(4x)dx The first step in evaluating this integral is to apply integration by parts: ∫udv = uv−∫vdu where u = and dv=h(x)dx where h(x) = Note: Use arcsin⁡(x) for sin−1⁡(x). After integrating by parts, we obtain the integral ∫1/4 0 vdu=∫1/4 0 f(x)dx on the right hand side where f(x) = The most appropriate substitution to simplify this integral is x=g(t) where g(t) = Note: We are using t as variable for angles instead of θ, since there is no standard way to type θ on a computer keyboard. After making this substitution and simplifying (using trig identities), we obtain the integral ∫b ak(t)dt where k(t) = a = b = After evaluating this integral and plugging back into the integration by parts formula we obtain: ∫1/4 0 xsin^−1⁡(4x)dx =

u = sin^-1(4x) h(x) = x f(x) = 4x^2/ (2)sqrt(1-16x^2) g(t)= sin(t)/4 k(t) = sin^2(t)/4 a = 0 b = 1.571 ∫1/4 0 xsin^−1⁡(4x)dx = pi/128

Evaluate the indefinite integral. ∫ln⁡(x^2+12x+35)dx Answer =

xln(x^2+12x+35)+6ln(x^2+12x+35)+ln|x+7|−ln|x+5|−2x−12 + C

Evaluate the indefinite integral. ∫[ln⁡(2x)]^2 dx =

xln^2(2x)−2(xln(2x)−x) +C

Use integration by parts to evaluate the integral. ∫(ln⁡(4x))^2 dx=

xln^2(4x)−2(xln(4x)−x) + C

Given the equation: xy=16, set up an integral to find the length of path from x=a to x=b and enter the integrand below. (i.e. if your integral is L=∫b a (c^2x^2)/h dx enter ((c^2x^2)/h) as your answer.) L=∫b a ____ dx

xy=16⇒y=16/x⇒dy/dx=−16/x^2 Thus L=∫b a sqrt(1+(dy/dx)^2)dx= ∫b a sqrt(1+(−16/x^2)^2)dx= ∫b a sqrt(1+(256/x^4))dx

∫2ytan^−1⁡(8y)dy = Use arctan⁡() to denote tan−1⁡() in your answer.

y^2arctan(8y)+(1/64)arctan(8y)−(1/8)y +C

Find the inverse function to y=f(x)=7sqrt(x+5). x=g(y)=

y^7-5

Evaluate the indefinite integral. ∫1/(x^2+x−6)dx =

−(1/5)(ln|(6/5)+(2/5)x|−ln|−(4/5)+(2/5)x|) +C.

Find the integral. ∫e^(2x)sin⁡(7x)dx=

−(7e^(2x)cos(7x)/53)+(2e^(2x)sin(7x)/53 ) + C

Verify that 1/(x^2−1) = (1/2)((1/x−1)−(1/x+1)) and use this equation to evaluate ∫3 2 1/(x^2−1)dx

−(ln(2)−ln(3))/2

Evaluate the definite integral. ∫5 1 t^2ln⁡(4t)dt=

−124/9+(ln(2^(248)·5^(125))/3)

Evaluate the definite integral. ∫π/12 −π/14. tan^2⁡(3x)dx

−13π84+13(1+tan(3π14))

Evaluate the limit using L'Hopital's rule lim x→π/2 7cos⁡(−7x)sec⁡(−9x)

−49/9

Evaluate the definite integral. ∫e 1 −9t^5ln⁡(t) dt Answer =

−5e^6+1/4

Let f(x)=tan^−1⁡(cos⁡(5x)) f′(x)=

−5sin(5x)/cos^2(5x)+1

Evaluate the indefinite integral. ∫x^2/(49+x^2)dx =

−7arctan(x/7)+x +C.

Find the value of ∫π/2 0 cos⁡(x)sin⁡(sin⁡(x))dx.

−cos(1)+1

Evaluate the indefinite integral. ∫tan^2⁡x dx Answer:

−x+tan(x) + C

Evaluate the integral ∫dx/x^2sqrt(36−x^2) using trigonometric substitution.

−√(36−x^2)/36x+C

Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If it diverges to infinity, enter inf If it diverges to negative infinity, enter -inf If it diverges without being infinity or negative infinity, enter div ∫∞ 3 ln⁡(2x)/x dx=

∞

If ∞ y=∑ (k+1)x^(k+3) k=0 then y′=

∞ ∑ (k+1)(k+3)x^(k+2) k=0

Consider the following indefinite integral. ∫(6x^3+2x^2−46x−12)/(x^2−9) dx The integrand decomposes into the form: ax+b+(c/x−3)+(d/x+3) Compute the coefficients: a= b= c= d= Now integrate term by term to evaluate the integral. Answer:

∫(6x^3+2x^2−46x−12)/(x^2−9)dx = ∫(6x+2+(5/x−3)+(3/x+3))dx a= 6 b= 2 c= 5 d= 3 Answer: 3x^2+2x+4ln|x^2−9|−ln|(x/3)+1|+ln|(x/3)−1|−27 + C

f(x)=(x+4)/(x+7) f^−1(−3)=

-6.25

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. y=x^2−15x+56, y=0; about the x-axis. volume =

0.10472

Referring to the figure above, find the volume generated by rotating the region R1 about the line AB. Volume =

0.314

∫e^4 e (ln⁡(s))/2s. ds=

15/4

Using disks or washers, find the volume of the solid obtained by rotating the region bounded by the curves y=x^2, x=5, and y=0 about the x-axis. volume =

1963.495

If f(x)=x^2, x≥0, then f^−1(5)=

2.2361

Find the volume of the solid formed by rotating the region enclosed by x=0, x=1, y=0, y=8+x^8 about the x-axis.

206.8318

Suppose that f(x)=log8⁡(3x^(2)+1).Find f′(2). f′(2) =

4/(13ln(2))

The inverse of the function f is (A, B, C or D):

Rewrite the expression ln⁡(a+b)+3ln⁡(a−b)−4ln⁡c as a single logarithm ln⁡A. Then the function A=

A = ((a+b)(a-b)^3)/c^4

Use the Laws of logarithms to rewrite the expression log2⁡(x^11y^11) in a form with no logarithm of a product, quotient or power. After rewriting we have log2⁡(x^11y^11)=Alog2⁡x+Blog2⁡y with the constant A= and the constant B=

A = 11 B = 11

Use the Laws of logarithms to rewrite the expression ln⁡((x^12)sqrt(y^14/z^18)) in a form with no logarithm of a product, quotient or power. After rewriting we have ln⁡((x^12)sqrt(y^14/z^18))=Aln⁡x+Bln⁡y+Cln⁡z with the constant A= the constant B= and the constant C=

A = 12 B = 7 C = -9

Use the Laws of logarithms to rewrite the expression log⁡((x^8y^10)/z^6) in a form with no logarithm of a product, quotient or power. After rewriting we have log⁡((x^8y^10)/z^6)=Alog⁡(x)+Blog⁡(y)+Clog⁡(z) with A= B= and C=

A = 8 B = 10 C = -6

Rewrite the expression ln⁡9+2ln⁡x+2ln⁡(x^2+2) as a single logarithm ln⁡A. Then the function A=

A = 9x^2(x^2+2)^2

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the curves y=2+x−x^2 and y+x=2 about the y-axis. Below is a graph of the bounded region. volume =

The line y+x=2 intersects the parabola y=2+x−x^2 when 2−x=2+x−x^2⟹x^2−2x=0⟹x(x−2)=0, that is, when x=0 or x=2. The region is rotated around the y−axis, so we have thin vertical cylindrical shells, of thickness dx. The interval needed by x to fill the region bounded by the curves is 0≤x≤2. For the integrand, we think of a cylindrical shell at x with height h=(2+x−x^2)−(2−x) =2x−x^2, thickness dx, rotated around the line x=0 and so has circumference 2πr=2πx. The volume then is V=∫2 0. 2πx(2x−x^2)dx =2π∫2 0(2x^2−x^3)dx =2π[2*x^3/3−x^4/4]2 0 =8/3π(cubic units)

Find the volume of the solid formed by rotating the region inside the first quadrant enclosed by y=x^4,y=64x about the x-axis. Volume =

The region to be rotated about the x axis is shown below. The curve y=x^4 intersects the line y=64x, when x^4=64x⟹x(x^3−64)=0, that is, when x=0 or x=4. We use the method of slicing, that is, V=∫b aA(x)dx, with each slice a washer with thickness dx. The area of each washer slice is A(x)=π(R^2−r^2) where each radius depends on the value of x, and the range of x for the region is 0≤x≤4. The largest radius, R, goes from the x−axis to the line y=64x , and so is R=64x. The smallest radius, r, goes from the x−axis to the curve y=x^4 , and so is r=x^4. So V=∫b aA(x)dx =∫b aπ(R^2−r^2)dx =∫4 0 π[(64x)^2 (x^4)^2]dx =∫4 0. π[4096x^2−x^8] dx =π[4096*x^3/3−x^9/9]4 0 = 524288/9π

Using disks or washers, find the volume of the solid obtained by rotating the region bounded by the curves y=x^2, 0≤x≤5, y=25, and x=0 about the y-axis. Volume =

The region to be rotated about the y-axis is shown below. The curve x=sqrt(y) is the right half of the parabola y=x^2. We use the method of slicing, that is, V=∫d cA(y)dy, with each slice a disk with thickness dy. The area of each disk slice is A(y)=πR^2, where the radius depends on the value of y, and the range of y for the region is 0≤y≤25. The radius R goes from the y-axis (x=0) to the curve x=sqrt(y). So R=sqrt(y)−0=sqrt(y), and V=∫d cA(y)dy =∫d cπR^2dy =∫25 0π(sqrt(y))^2dy =∫25 0π(y)dy =π[y^2/2]25 0 =π[25^2/2] =625/2π (cubic units).

Using disks or washers, find the volume of the solid obtained by rotating the region bounded by the curves x=y−y^2 and x=0 about the y-axis. Volume =

The region to be rotated about the y-axis is shown below. The curve x=y−y^2 intersects the vertical line x=0 at y=0 and y=1. We use the method of slicing, that is, V=∫d cA(y)dy, with each slice a disk with thickness dy. The area of each disk slice is A(y)=πR^2, where the radius depends on the value of y , and the range of y for the region is 0≤y≤1. The radius R goes from the y-axis (x=0) to the curve x=y−y^2. So R=y−y^2−0=y−y^2, and V=∫d cA(y)dy =∫dcπR^2dy =∫1 0π(y−y^2)^2dy =∫1 0π(y^2−2y^3+y^4)dy =π[y^3/3−y^4/2+y^5/5]1 0 =π[1/3−1/2+1/5] =π/30 (cubic units).

Find the volume of the solid obtained by rotating the region under the graph of the function f(x)=x^9/8 about the x-axis over the interval [1,3]. V=

The volume V of the solid of revolution is V=π∫3 1. (x^9/8)^2dx =π∫3 1x^9/4dx =4π/13*x13/4 | 3 1 =4π/13[3^13/4−1] =33.3820786198627

Find the volume of the solid obtained by rotating the region under the graph of the function f(x)=2x+1 about the x-axis over the interval [0,1]. V =

The volume of the solid of revolution is π∫1 0. (2x+1)^2dx = 4π∫1 0. (x+1)^−2dx =−4π(x+1)^−1|1 0 =2π

Use the Shell Method to find the volume of the solid obtained by rotating region under the graph of f(x)=x^2+2 for 0≤x≤5 about the y-axis. V =

When rotating the region in question about the y-axis, each shell has radius x and height x^2+2. The volume of the resulting solid is 2π∫5 0. x(x^2+2)dx =2π∫5 0(x^3+2x)dx =2π(1/4*x^4+x^2)|5 0 =362.5π

The graph of the function f(x)=log4⁡(x−3) can be obtained from the graph of g(x)=log4⁡(x) by one of the following actions: (a) shifting the graph of g(x) to the right 3 units; (b) shifting the graph of g(x) to the left 3 units; (c) shifting the graph of g(x) upward 3 units; (d) shifting the graph of g(x) downward 3 units; Your answer is (input a, b, c, or d) The domain of the function f(x) is . Note: Enter your answer using interval notation. The range of the function f(x) is . Note: Enter your answer using interval notation. The x-intercept of the function f(x) is . The vertical asymptote of the function f(x) has equation:

Your answer is (input a, b, c, or d) = a The domain of f(x) is =(3,inf) The range of f(x) is = (-inf,inf) The x-intercept of f(x) is = (4,0) The vertical asymptote of f(x) has equation = No correct answer specified

If ln⁡(a)=2, ln⁡(b)=3, and ln⁡(c)=5, evaluate the following: (a) ln⁡(a^2/(b^1c^1))= (b) lnsqrt(⁡b^−2c^−4a^−4)= (c) ln⁡(a^3b^−3)/ln⁡((bc)^−1)= (d) (ln⁡ c^2)(ln⁡a/b^−1)^2=

a) -4 b) -17 c) 0.375 d) 250

Find the indicated derivatives. (a) d/dx [e^(x^5)+log3⁡(π)] = (b) d/dx [(sqrt^7(x))^ln⁡(x)] =

a) 5e^(x^5)x^4 b) 2e^(ln^(2)(x)/7l)n(x)/7x

Evaluate the following expressions. Your answers must be exact and in simplest form. (a) ln⁡e^9= (b) e^ln⁡4= (c) e^ln⁡sqrt(4)= (d) ln⁡(1/e^4)=

a) 9 b) 4 c) 2 d) -4

Suppose f(x)=2x−7 and g(y)=y/2+7/2. (a) Find the composition g(f(x))= (b) Find the composition f(g(y))= (c) Are the functions f and g inverse to each other?

a)x b)y c)yes

A function f(x) is graphed in plane A. It is clearly a 1:1 function, so it must have an inverse.Enter the color ("red", "green", "blue", or "yellow") of this inverse function which is graphed in plane B. Use what you know about the graphs of inverse functions rather than algebraic calculations based on what you might guess the function to be. Color of f^−1 graph =

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