Final Exam

Question 1

For trig expressions, when the denominator is 3 or 6, use the numbers for the triangle

Answer 1

1, 2(hyp), sqrt 3 (opp)

Question 2

For trig expressions whose denominator is 2 or 4

Answer 2

Use the numbers 1, 1, sqrt 2 (hypt)

Question 3

Chain rule

Answer 3

Multiply function by exponent then subtract the exponent by 1, then multiply the expression by the derivative of the inside

Question 4

Pythagorean, identities of sine and cosine

Answer 4

Sine squared X plus cosine squared X equals one

Question 5

Integration by parts

Answer 5

U times V minus the integral of V times DU

Where u is one part, du is derivative of u, dv is the other part, and v is the integral of dv

To make the table: set up a list of derivates of u that alternate between positive and negative until you get to 0, and next to it make a list of the integrals of dv (the other part). The answer is diagnollay multiplying and adding the next part.

Question 6

For integrals with trigonometric substitution, the steps are

Answer 6

Write what x and dx equals
Substitute the x’s and dx
Simplify the radical by factoring and seeing if you get any identities and take the sqrt
See if anything cancels

Question 7

How to find area between two curves

Answer 7

1) Graph the functions
2) Find the intersection point by setting both functions equal to each other
3) Set up the integral from a to b (a being the left bound placed on bottom of integral and b being the right bound placed on top of the integral)
4) The integral inside is the top function minus the bottom function dx
5) Integrate and then plug in b and subtract plugged in a

Question 8

How to find area between two curves in terms of y

Answer 8

1) Rewrite the function from y= to x=
2) Find the intersection point by setting both functions equal to each other
3) Set up the integral from a to b with a being the lowest bound on the bottom of integral and b being the highest bound on the top of the integral
4) The integral inside is the right function minus the left function dy
5) integrate and then plug in b and subtract plugged in a

Question 9

How to find volume w/ cross sections (semicircle)

Answer 9

integral from a to b 1/2*pi((f(x)-g(x))/2)^2 dx
= 1/2pi int from a to b ((f(x)-g(x))/2)^2 dx

Question 10

How to find volume w/ cross sections (squares)

Answer 10

int from a to b (f(x)-g(x))^2 dx

Question 11

How to find volume w/ cross sections (rectangle)

Answer 11

int from a to b k*(f(x)-g(x))^2 dx

(height is k of length)

Question 12

How to find volume w/ cross sections (circle)

Answer 12

int from a to b pi(1/2(f(x)-g(x))^2 dx

Question 13

How to find volume w/ cross sections (isosceles right triangle)

Answer 13

int from a to b 1/2(f(x)-g(x))^2 dx

Question 14

How to find volume w/ cross sections (equilateral triangle)

Answer 14

int from a to b (sqrt of 3)/2*(f(x)-g(x))^2 dx

Question 15

Disk method

Answer 15

int from a to b pi(f(x))^2 dx

Question 16

Washer method

Answer 16

int from a to b pi[(R(x))^2-(r(x))^2] dx

big R is the outer radius (or top function)

Question 17

Shell method

Answer 17

int from a to b 2pixf(x) dx

Question 18

Arc length

Answer 18

int from a to b sqrt: 1+(f’)^2 dx

Question 19

Surface area

Answer 19

int from a to b 2pif(x)sqrt: 1+(f’)^2 dx

Question 20

Find work of movement

Answer 20

int from 0 to h (weight) dx

Question 21

Find work of spring

Answer 21

int from a to b k*x dx

F=kx
F= force, x=distance adjusted

Question 22

Form of geometric series and when it converges/diverges

Answer 22

a * r^n-1

|r|< 1 : converges
|r| ≥ 1 : diverges

Question 23

Sum of geometric series formula

Answer 23

S = a/(1-r)

Question 24

P-series convergence/divergence

Answer 24

P<1 : diverge
P ≥ 1 : converge

Question 25

Direct comparison test steps

Answer 25

Determine, as n grows, what pieces will be insignificant (remove additional term such as +2 in the series 1/n+2.) This will be what you compare to. (1/n in the example) First compare to determine if the series you’re comparing to is bigger/smaller. Then determine if bn (bigger function) converges or diverges by using tests like geometric or p-series. If bn converges, an converges as well. If bn diverges, an is inconclusive & you have to use the limit comparison test. If an diverges, bn diverges An: smaller series Bn: bigger series

Question 26

Limit comparison steps

Answer 26

-Divide the series you’re comparing to by the series you’re given. Ex: Given: 1/n+10 Series to compare: 1/n Therefore, (1/n)/(1/n+10) -rewrite into multiple (KCF) 1/n x (n+10)/1 -cancel terms (n+10/n) -evaluate limit =1 (nth term test) Therefore, diverge

Question 27

Alternating series test

Answer 27

If the limit of the series without the alternating piece goes to 0, the series converges If it doesn’t go to 0, it diverges by the nth term test

Question 28

Absolute convergence & conditional convergence

Answer 28

By using AST, determine if alternating series converges or diverges. If the series converges, and, without its alternating piece, converges, then the series is absolutely convergent If the series converges, and, without its alternating piece, diverges, then the series is conditionally convergent If the series diverges by failing the AST, then the series diverges by the nth term test

Question 29

Ratio test

Answer 29

Plug in n+1 for n in the series to get the series an+1 Multiply an+1 by the reciprocal of the original series Cancel terms Evaluate limit -L=limit -0 ≤ L < 1 ; converge absolutely -L > 1 ; diverges -L = 1 ; ratio test inconclusive

Question 30

Root test

Answer 30

In a series that is raised to power n, Find the limit of the series to the n root This gets rid of the power n Evaluate the limit -L=limit -0 ≤ L < 1 ; converge absolutely -L > 1 ; diverges -L = 1 ; ratio test inconclusive

Question 31

Radius and Interval of Convergence

Answer 31

Use the ratio test Set the absolute value limit less than 1 The radius of convergence is the limit with the absolute value bars less than R. R is the radius of convergence. ex: 1/9|x-4| < 1 |x-4| < 9 The radius of convergence is 9. Find the interval of convergence by setting the limit between the negative and positive value you got from the radius of convergence. Solve by isolating x in the middle. ex: -9 < |x-4| < 9 -5 < x < 13 Test the endpoint values by plugging them in for x in the series. If div; not () If conv; included []

Question 32

Taylor Polynomial Form + Listed out to P2

Answer 32

(f^n(a)/n!)(x-a)^n f(a)+f’(a)(x-a)+(f’‘(a)/2!)(x-a)^2

Question 33

Finding Taylor Polynomials

Answer 33

List out the f(a)’s P0 is “a” (which is the x value given) plugged into f(x). P1 is “a” (which is the x value given) plugged into the first derivative of f(x). P2 is “a” (which is the x value given) plugged into the 2nd derivative of f(x). etc.. Now use the Taylor polynomial form and list out the taylor polynomial.

Question 34

Taylor Polynomials: Use these polynomials to estimate f(x) where x is a given number.

Answer 34

Plug in whatever x was replaced by into the polynomials and solve each polynomial.

Question 35

Taylor Polynomials: Use Taylor’s Remainder Theorem to bound the error.

Answer 35

1) Whatever the highest derivative to find the polynomial was, find the next derivative after that. ex: if the highest polynomial found was P2, find what the next derivative would be. Basically, what you would plug in for P3. Derivative for P2 was (-2/9)x^-5/3 Derivative for P3 is (10/27)x^-8/13 2) Then plug in the x= number that was given into the new derivative. (10/27)(8)^-8/13 = .00145 3) Then find the next Taylor polynomial value (the numerator is the value from step 2) and plug in x from what x was replaced by when asked to estimate. (.00145/3!)(11-8)^3

Question 36

AST Error Bound

Answer 36

|Rn| ≤ bn+1 The partial sum is only off from the true sum by less than the next term The answer will be the next term after the partial sum. Ex: Given the AST: ((-1)^n+1)/n^2, determine a bound on the error R10 if we approximate the sum of the series by the partial sum S10. |R10| ≤ |a11| |R10| ≤ |1/11^2| |R10| ≤ 1/121

Question 37

Integral Test Error Bound

Answer 37

Set up an integral with the bounds from the value you’re testing to infinity. ex: For the series: 1/n^4, calculate S5 and estimate the error R5 integral from 5 to infinity: 1/x^4 dx Solve the integral and plug in the bounds and solve. integral from 5 to infinity: 1/x^4 dx =-1/3x^3 from 5 to infinity =0+1/3(5)^3= 0.0027

Question 38

Finding the area between two polar curves

Answer 38

integral from alpha to beta: 1/2 [(r1)^2-(r2)^2]dθ

Question 39

Finding the length of a polar curve

Answer 39

integral from alpha to beta: sqrt of r^2+r'^2

Question 40

Expressing the following integration as an infinite series steps

Answer 40

Determine what series you have that matches with your integral. Using the maclaurin series that matches to your integral, make the same modification to your series that your integral has. ex: int: cos(x^2) dx compare it to cosx= (-1)^n * (x^2n)/(2n)! Make the same modifications to the comparing series that's in your integral (replace x with x^2) cos(x^2)= (-1)^n * ((x^2)^2n)/(2n)!