MATH 19B study guide review reimen sums and maybe 5.9/5.1 also look up tricky integral solutions like 1/a root fix 35

power reduction of cos^2(a)

1+cos2a/2

power reduction of sin^2(a)

1-cos2a/2

6.2Average value = the average value of an integrable function f on [a,b] is the quantity

1/b-a integral f(x) dx

let f(x)=integral 0->2x(3sin(t) dt (a)f'(pi/4) (b) compute f''(x)

3sin(2x)*2 because of chain rule then substitute =6 b) 24sin(2x)cos(2x)

WHAT IS THE INTEGRAL A tank has ends that are semicircles 4 meters across at the top and is 18 meters long. The tank is full of water. Find the work done in pumping the water just over the top of the tank. State the units of your answer. Recall that the density of water is 1000 kg/m . $

A=0 B=2 WATER AND GRAVITY AND 18* INTEGRAL (Y)(2*ROOT(4-Y^2))

INTEGRAL OF A tank has ends that are trapezoids 8 meters wide at the top and 6 meters wide at the bottom. The tank is 4 meters high and 10 meters long. The tank is filled with water to a depth of 3 meters. Find the work done in pumping the water just over the top of the tank. State the units of your answer. Recall that the density of water is 1000 kg/m .

GRAVITY AND WATERINTEGRAL0->3 (4-Y)10(6+1/2Y

LIATE

Log Inverse Algebraic Trig Exponential

6.2Total Mass for an integral

M=integral p(x) dx p(x)=linear mass density

LEFT MIDPOINT AND RIGHT ALL DEDEPD

ON THE GRAPH FOR WHIC IS BIGGER

5.1example of Approximating Area by Rectangles Calculate R4 and R6 for f (x) = x2 on the interval [1, 3].

Solution Step 1. Determine x and the right endpoints. To calculate R4, divide [1, 3] into four subintervals of width x = (3−1)/4= 1/2. The right endpoints are the numbers xj = a + jx = 1 + j(1/2) for j = 1, 2, 3, 4. They are spaced at intervals of 1/2 beginning at 3/2, so, as we see in Figure 5(A), the right endpoints are 3/2, 4/2, 5/2, 6/2 . Step 2. Calculate x times the sum of function values. answer 271/27

5.6Net Change as the Integral of a Rate of Change

The net change in s(t) over an interval [t1, t2] is given by the integral s'(t)dt=s(t2)-s(t1)

SOLVE XROOT 3X+5

U=3X+5 then U-5/3=X (U-5/3)(U^1/2) 1/9(U-5)(U^1/2) 1/9U^3/2-5U^1/2=1/9(2/5U^5/2-10/3 *U^3/2) THEN SUBSTITUTE AND DISTRIBUTE

HOW TO SOLVE THE INTEGRAL OF 0->6 5X-X^2 with reiman sums

X=6I/N SUMMATION (5((6I/N))-(6I/N)^2)triangle x =6/n THEN EXPAND AND SIMPLIFY

ASSUMING A SPRING CONSTANT OF K=400N/M FIND THE WORK REQUIRED To: (a) Stretch the spring 10 cm beyond equilibrium. (b) if at -0.03 Compress x-.05

a integral 0->.1 400x b) integral -.03->-.05 400x

5.1 Approximating Area by Rectangles

a = left endpoint of interval [a, b] b = right endpoint of interval [a, b] N = number of subintervals in [a, b] triangle x=((b-a)/n)*i x= +triangle x

5.8 integral of b^x=

b^x/(lnb)+c

5.8 Inverse Trig Functions integrals sin, tan,sec

cos-1(x) =-sin^-1(x) cot(x)=-tan^-1(x) csc^-1=-sec^-1(x)

5.7Change of Variables Formula for Definite Integrals

f(u(x)))u'(x)dx=[u(a),u(b)]f(u)du

6.5work=

force* distance

Integration by parts formula

ie reverse product rule f(x)g(x)- integral f'(x)g(x)dx where g(x)=integral of g(x)

5.3 thm 3 Antiderivative of y = 1/x The function F(x) = ln |x| is an antiderivative of y = 1/x in the domain {x : x = 0}; that is,

integral dx/x=ln|x|+C

To integrate an odd power of sin x times cosn x, write

integral sin^(2k+1)xcos^(n)xdx= integral(1-cos^2x)^k cos^nxsinx then use substitution u=cosx du=-sinx

To integrate an odd power of cos x times sinm x, write

integral sin^mxcos^(2k+1)xdx=integral (sin^mx)(1-sin2x)^k cosx dx then us substitution u=sin x,du=cos xdx

6.1area between the curves

integral(ytop-ybot) dx=integral (f(x)-g(x))dx if they change check and equal them each other

6.4 When finding a volume using the shell method, the shell height

is always parrallel to the axis of rotation when finding a volume using the Dish and Washer Method the disk radius is always perpendicular to the axis of rotation

5.6 distance

is the integral that that accounts for all distance traveled and negative distance therefore you have to split it so your going to take the absolute value of the integral

5.6 displacement

is the regular integral

5.7 The Substitution Method integrals If F(x) = f (x), then

it is basically the chain rule in reverse f(u(x))u'(x)dx=F(u(x))+C

5.1 power sums of summations

j=n^2/2+n/2 j^2=N^3/3+N^2/2+N/6 j^3=n^4/4+N^3/2+N^2/4

5.1 left right and midpoint approximation

left summation j=0->n-1 right summation j=1->n mid point j=0->n-1 (xj+xj+1)/2

5.4 the defintie integral of dx/x

ln|b|-ln|a|=ln(b/a)

how toReversing the Limits of Integration For a < b, we set

makes it negative

with trig substitution you want to

match the form and keep in the same boat of identities ie x^2+1=sectheta

6.2 total population

p=2pi integrals [0,R] rp(r) p(r)=radial density

x^2/(x^2+16)^3/2

pull out the 16 by square rooting the cubing make sure to translate the variable for these equation in terms of a substitution for this it will be dtheta

6.2laminar flow rate

q= 2pi integrals [0,R] rv(r) dr v(r)=velocity at radius r

5.2the defintie integral is the

signed area of region between the graph and x-axis over [a,b]

5.3basic trigonometric integrals and e^x

sinx=-cosx+c sec^2=tanx+c secxtanx=secx cosx=sinx+c csc^(2)x=-cotx+c cscxcotxdx=-cscx+c e^kx=1/k(e^(kx))+c

tips for trignometric substitution

step 1.substitute to elimninate the square root evaluate the trignometric integral convert back to orignal variable

7.3 trigonometric substitution

substituing our trig integrals for x to solve our integral for mostly roots.

how to solve 1/root(1-x^2)

substitute sin theta to eliminate the square root cos theta over cos theta

how to solve tan^2(x)sec^6(x)

tan^2(x)(tan^2(x)+1)*sec^4(x)dx u=tan(x) u^2(u^2+1)sec^4(x)dx u^2(u^2+1)sec^2(x) du u^2(u^2+1)u^2+1 u^2(u^4+2u^2+1) u^6+2u^4+u^2 1/7u^72/5u^5+1/3u^3+c then subsitute tan(X)

6.1 Area between the graphs

this is for changing the axis rotated integral (xringht-xleft)dy=(g(y)-h(y))dy

hoe to solve 3x^3e^3x^2

thsi is an integration by parts we say that the derivative is going to be x

6.2formulas volue

v= integral of equation A(y) dy A(y)=cross sectional area

6.4volume of revolution the shell method

v=2pi integral(radius)(height of shell)=2pi integral xf(x)

6.3 volumes of revolution disk method

v=pi integral(r^2outer-r^2inner)dx=pi integral(f(x)^2-g(x)^2)dx

5.2 THEOREM 4 Additivity for Adjacent Intervals Let a ≤ b ≤ c, and assume that f is integrable. Then

you can split integrals into seprate integrals as long as the bounds are connecting

5.1Linearity of Summations

you can split summations you can pull constants out and then multiply them also summation C=nc where c is any constant and n>=1

5.2 the definite integral and rules

you can take out constants you can split integrals if added