Calculus 2 final

cos^2x (DOUBLE ANGLE)

(1+cos2x)2

sin^2x (DOUBLE ANGLE)

(1-cos2x)/2

derivative for parametric equations

(dy/dt)/(dx/dt)

Find (f^-1)'(5) if f(x)= e^x+3x+4

(f^-1)'(x)=1/f'(x)=(1/4) 5=e^x+3x+4 x=0

derivative of arccosx

-1/squareroot(1-x^2)

derivative of csc(x)

-csc(x)cot(x)

derivative of cot(x)

-csc^2(x)

derivative of cos(x)

-sin(x)

sin^2x+cos^2x=

1

derivative of arctanx

1/(1+x^2)

does the series 1/5^n sum from n=2 to infinity converge or diverge, explain

1/(1-(1/5)) - 1 - (1/5) = 1/20 converges (GEOMETRIC r=1/5)

(f^-1)'(6) if f(x)=5+3x+e^x

1/(3+e^0)=(1/4) 6=5+3x+e^x

(f^-1)'(x)

1/f'(x)

derivative of arcsinx

1/squareroot(1-x^2)

integral of ((1-y)/y)^2 dy

= integral of (1-2y+y^2/ y^2) dy = -1/y-2ln lyl +y+C

partial fraction do not solve for coefficients for (x+5)^2/((x^2+1)^2(x^2-1)^2)

A/(x+1)+B/(x+1)^2+C/(x-1)+D/(x-1)^2+(Ex+F)/(x^2+1)+(Gx+H)/(x^2+1)^2

write out the partial fraction do not solve for numerical values for x/((x+2)^2(x-3))

A/(x-3)+B/(x+2)+C/(x+2)^2

find integral of (x+9)/(x(x-3)^2) dx using partial fractions

A/x+B/(x-3)+C/(x-3)^2 when x=3 C=4 when x=1 A=(6+2B)/4 when x=-1 B=-1 A=1 integral 1/x dx - integral 1/(x-3)dx +4 integral 1/(x-3)^2 dx =ln lxl - ln lx-3l - 4/(x-3) + C

write out the partial fraction do not solve for the numerical values for 5x^2/((x^3(x+1)^3)

A/x+B/x^2+C/x^3+D/(x+1)+E/(x+1)^2+F/(x+1)^3

sum from n=1 to infinity of (2^n n!)/(n+2)! the series converges or diverges, test used, work

Diverges Ratio Test L= lim n to infinity of (2^(n+1)(n+1)!)/(n+3)!)/((2^n)(n!)/(n+2)!) = lim n to infinity of 2(n+1)/(n+3)=2>1=L the series converges

arc length equation

L integral from a to b of squareroot( (dx/dt)^2+(dy/dt)^2) dt

find the length of the curve given y=ln(cosx), 0<=x<=pie/3

L= integral from 0 to pie/3 squareroot(1+(-sinx/cosx)^2) dx = ln lsecx +tan xl from 0 to pie/3= ln(2+squareroot3)

lim x to 0 of (1-a^x)/(1-b^x)

LH (0/0)= lim x to 0 (-a^x lna)/ (-b^x ln b) = lna/lnb

find the limit lim x to 0 of e^x -1-x/(7x^2)

LH (0/0)=lim x to 0 e^x -0-x/14x = LH (0/0)= lim x to 0 e^x/14= 1/14

lim x to infinity of x^2 e^-x

LH (infin./infin)= lim x to infinity 2x/e^x LH= lim x to infinity 2/e^x = 0

1-(1/2)+(1/5)-(1/10)+(1/17)-(1/26)+(1/37)+... the series converges or diverges, test used, work

converges Alternating Series Test sum from n=1 to infinity (-1)/(n^2+1) (1/2)>(1/5)>(1/10)>(1/17)>(1/26)>(1/37)... and lim n to infinity of 1/(n^2+1)=0

sum from n=1 to infinity of arctan(n)/n^5, the series converges or diverges, test used, work

converges Comparison Test sum from n=1 to infinity of (pie/2)/n^5, convergent p-series 5>1 so sum from n=1 to infinity of arctan(n)/n^5 converges too

sum from n=1 to infinity of ne^(-n^2), the series converges, or diverges, test used, work

converges Integral Test integral of xe^(-x^2) from 1 to infinity converges so the sum does too

(1/2)+(1/5)+(1/10)+(1/17)+(1/26)+(1/37)+... the series converges or diverges, test used, work

converges Integral Test sum from n=1 to infinity of 1/(n^2+1), integral of 1/(x^2+1) dx from 1 to infinity=pie/4, integral converges so the sum converges too

sum from n=2 to infinity of the ((nth root of e) -1)^n, the series converges or diverges, test used, work

converges root test L=lim n to infinity l ((nth root of e) -1)^n l^(1/n)= 1-1=0<1 converges by the root test

derivative of sin(x)

cos(x)

1+cot^2x=

csc^2x

sum from n=1 to infinity of (n^2+5)/(squareroot(n^6+n^5+1), the series converges or diverges, test used, work

diverges Limit Comparison Test an is about 1/n so sum from n=1 to infinity of 1/n p-series p=1 so diverges, C=lim n to infinity of ((n^2+5)/(squareroot(n^6+n^5+1))/(1/n)=1 not 0 not infinity so both series diverge

differentiate arctan (arcsin squareroot x)

f'(x)=1/(1+(arcsin squareroot x)^2)(1/squareroot(1-(squareroot x)^2)) (1/2squarerootx) = 1/2 (1/squareroot(x(1-x))(1+(arcsin (squareroot x)^2))

indefiiate integral of 1/squareroot (x^2-4) dx

integral 1/squareroot(4sec^2theta-4) times 2sectheta tantheta dtheta= integral sectheta dtheta= ln lsectheta +tanthetal +C (draw triangle) = ln lx/2+ squareroot(x^2-4) /2l +C

integral from 0 to pie/4 of (1-tan^2theta)/(sec^2theta) dtheta

integral from 0 to pie/4 (1- (sin^2 theta/cos^2 theta)) cos^2theta dtheta = integral from 0 to pie/4 of cos 2theta (USED DOUBLE ANGLE)= 1/2

does the sequence arctan((2-n^3)/(1+n^2)) fron n=2 to infinity converge or diverge, explain

lim n to infinity arctan ((2/n^3)-1)/((1/n^3)+(1/n))=-pie/2

does the sequence (6^n+4^n)/2^n n=1 to infinity converge or diverge, explain

lim n to infinity of (6^n+4^n)/(2^n) = lim n to infinity (6/2)^n+(4/2)^n = infinity so diverges because the limit of the sum approaches infinity

does the series (2^n - 2^(n+1)) Sum from n=1 to infinity converge or diverge, explain

lim n to infinity of 2^n+1= infinity diverges because you are subtracting a number that is getting closer to -infinity

find the limit lim x to 1^+ of (x-1)tan (pie/x times x)

lim x to 1^+ (x-1)/(cot(pie/2 times x)) LH (0/0) = lim x to 1^+ 1/((-pie/2)csc^2(pie/2 times x))= -2/pie

integral of cscx dx

ln lcscx-cotxl +C

integral of secx dx

ln lsecx+tanxl +C

integral of tanx dx

ln lsecxl +C

integral of cotx dx

ln lsinxl +C

lim x to pie/2+ of tanx^cosx

ln y= cosx lntanx lim x to pie/2+ lny= lim x to pie/2+ cosx lntanx= lim x to pie/2+ lntanx/secs (infin/infin) LH= lim x to pie/2+ secx/tan^2x= lim x to pie/2+ cosx/sin^2x=0 so lim x to 0 y=e^0=1

differentiate tanx^cosx

lny=cosxlntanx y'=(tanx^cosx)(-sinx(lntanx)+cscx)

differentiate the function y=x^(e^2x)

lny=e^(2x)lnx y'=e^(2x)(x^e(2x))(2lnx+(1/x))

indfinite integral (x^3-3x^2+18)/(x^3-3x^2) dx

remainder of 18 after division= integral 1+ 18/(x^3-3x^2) dx, do partial fractions= integral 1 dx+ integral -2/x dx + integral -6/x^2 dx + integral 2/(x-3)= x-2ln lxl +6(1/x)+2ln lx-3l +C

set up an integral for the surface area of a solid of a revolution around the curve x=t-t^2, y=(2/3)t^(3/2), 1<=t<=2 about the x-axis

s= integral from 1 to 2 of 2pie (2/3t^(3/2)) squareroot ((1-2t)^2+(t^(1/2))^2) dt

surface area about x-axis equation

s= integral from a to b of 2pie y(t) squareroot ((dx/dt)^2+(dy/dt)^2) dt

derivative of sec(x)

sec(x)tan(x)

derivative of tan(x)

sec^2(x)

1+tan^2x=

sec^2x

Integral of (e^(1/x))/(x^2)

u=(1/x) du=-(1/x^2) =-e^(1/x)+C

determine if the integral from -infinty to 0 of xe^(-x^2) dx is convergent or divergent. evaluate if convergent

u=-x^2 du=-2xdx -1/2 integral from -infinity to 0 of e^u du lim t to -infinity of -1/2e^u from t to 0 limit converges to -1/2

does the integral from 1 to infinity of ye^(-y^2) dy converge or diverge, evaluate if converges

u=-y^2 du=-2ydy -1/2 integral from -1 to -infinity e^u du= lim t to -infinity (1/2 e^u) from t to -1, converges to 1/2e^-1

integral of dx/x(1+lnx)

u=1+lnx du=1/x dx integral of 1/u du = ln lul +C = ln l1+lnxl +C

integral arctanx dx

u=arctanx du=1/(1+x^2) dx v=x dv=dx xarctanx- integral x/(1+x^2) dx (w=1+x^2) =xarctanx-1/2ln l1+x^2l +C

integral from 0 to pie/2 sin^3xcos^2x dx

u=cosx du=-sinx integral from 1 to 0 of -1u^2(1-u^2)du=2/15

integral from 0 to 2 of e^x/(1+e^2x) dx

u=e^x du=e^x dx integral from 1 to e^2 of 1/(1+u^2) du arctanu from 1 to e^2 = arctane^2- arctan1

integral of e^x tan^2(e^x) dx

u=e^x du=e^x dx integral tan^2u dx =integral (sec^2u -1) du= tan(e^x)-e^x+C

integral sin^3t cos^3t dt

u=sint integral u^3 (1-u^2)du= 1/4sin^4t-1/6sin^6t+C

integral of costheta/ (sin^2theta +sintheta) dtheta

u=sintheta du=costheta dtheta integral 1/(u^2+u) du (use partial fractions) integral of 1/u + integral of -1/u+1 du= ln lsinthetal -ln lsintheta+1l +C

evaluate the integral from 0 to (pie/2) of 2cosx/(1+sin^2x) dx

u=sinx du=cos(x)dx 2integral (1/1+u^2)du from 0 to 1 = pie/2

integral of xsecx tanx dx

u=x du=dx v=secx dv=secxtanxdx =secx-integral secx dx =xsecx-ln lsecx+tanxl +C

integral of xsec^2x dx

u=x v=tanx du=dx dv=sec^2x dx integral xsec^2xdx=xtanx-integral tanxdx =xtanx-ln lsecxl +C

does the integral from -2 to 0 of 1/(x+2)^2 dx converge or diverge, evaluate if converges

u=x+2 du=dx integral from 0 to 2 1/u^2 du, lim t to 0+ of -1/u from t to 2= DNE so divergent

the integral of x/(squareroot (25-x^2)) dx

x=5sintheta dx=5costhetat dtheta integral (5sintheta/squareroot(25-(5sintheta)^2))(5costheta dtheta)= 5 integral sintheta dtheta= -5costheta +C

squareroot (x^2-a^2)

x=asectheta

squareroot (a^2-x^2)

x=asintheta

squareroot (a^2+x^2)

x=atantheta

differentiate y = cos (e^x)+e^cosx

y'=-e^xsin(e^x)-e^cosx sinx

dy/dx if y=sin^2(e^cscx)

y'=2(sin(e^cscx))(cos(e^cscx))(e^cscx)(-cscxcotx)

equation for the tangent line to the curve y=3arccos(x/2) at the point (1,pie)

y'=m=(-3/(squareroot(1-(x/2)^2))(1/2)=-3/squareroot3 y-pie=(-3/sqareroot3)(x-1)

find the limit lim x to 0^+ of (4x+1)^cotx

y= lim x to 0^+ of (4x+1)^cotx ln y = lim x to 0^+ of ln((4x+1)^cotx) = lim x to 0+ (cot x) ln(4x + 1) = lim x to 0+ (ln (4x +1))/(tan x)) (0/0) = lim x to 0+ (4/(4x+1))/(sec2 x) = lim x to 0+ (4/(4x+1) sec2 x)= 4

differentiate y=ln(x^4 cos^2x)

y=4lnx+2lncosx y'=4/x+(2/cosx)(-sinx)= 4/x-2tanx

If a bacteria population starts with 100 bacteria and doubles every three hours, then the number of bacteria after t hours is n= f(t)= 100(2^(t.3)). When will the population reach 50000

y=Ce^(kt) f(t)=50000 50000=100(2^(t/3)) 500= 2^(t/3) ln 500= t/3 ln2 t=3ln500/(ln2)