Math 211

These problems are provided to help you study. The presence of a problem on this sheet does not imply that a similar problem will appear on the test. And the absence of a problem from this sheet does not imply that the test will not have a similar problem.

1. Compute .

2. Compute .

3. Compute .

4. Compute .

5. Compute .

6. Compute .

7. Compute .

8. Compute .

9. Compute .

10. Compute .

11. Compute .

12. Compute .

13. Compute .

14. Compute .

15. Compute .

16. Compute .

17. Compute .

18. Compute .

19. Compute .

20. Compute .

21. Compute .

22. Compute .

23. Compute .

24. Compute .

25. Compute .

26. Compute .

27. Compute .

28. Compute .

29. Compute .

30. How would you try to decompose using partial fractions? (Just write out the fractions --- you don't need to solve for the parameters.)

31. What is wrong with the following "partial fractions decomposition"?

32. What is wrong with the following "partial fractions decomposition"?

33. Find the partial fractions decomposition of

34. (a) Compute .

(b) Calvin Butterball tries to use the antiderivative from (a) to compute

He gets

Does this computation make sense? Why or why not?

35. Find the area of the region under from to .

36. Compute .

37. Compute .

38. Compute .

39. Compute .

40. Compute .

41. Compute .

42. Compute .

43. Prove that converges. [Hint: Compare the integral to .]

44. Prove that converges. [Hint: Use comparison, starting with the fact that .]

45. (a) Show that the following integrals both diverge:

(It follows that diverges as well.)

(b) Show that
converges. (This is called the * Cauchy principal
value* of the integral; this problem shows that is not the same as .)

1. Compute .

2. Compute .

3. Compute .

4. Compute .

5. Compute .

The second equality comes from dividing by (long division). Alternatively, you can do this:

6. Compute .

I'll do the antiderivative first:

Therefore,

7. Compute .

8. Compute .

9. Compute .

10. Compute .

11. Compute .

12. Compute .

13. Compute .

14. Compute .

15. Compute .

16. Compute .

17. Compute .

Since , I have

Set . This gives

Set . This gives

Therefore,

18. Compute .

Set . This gives

Set . This gives

Plugging the values for b and c back in yields

Set . This gives

Thus,

Hence,

19. Compute .

The top and the bottom both have degree 3, so I must divide the top by the bottom:

I'll put the 3 aside for now, and work on the fraction:

Clear denominators:

Set : I get , so .

Set : I get , so .

Plug B and C back in:

Differentiate:

Set : I , so .

Therefore,

Hence,

Finally,

20. Compute .

Since the least common multiple of 2, 3, and 4 is 12, I'll let :

21. Compute .

22. Compute .

Complete the square:

So

23. Compute .

Complete the square:

So

24. Compute .

Since and , I have

Therefore,

I did the first part of the u-integral using the substitution .

25. Compute .

First,

I note that and , so I needed 9 to complete the square.

Thus,

26. Compute .

I completed the square by noting that and . You can do the integral using .

27. Compute .

First, .

Set . I get , so .

Set . I get , so .

Then

At this point, you can plug other numbers in for x, or differentiate the equation and then plug numbers in. The idea is to get equations for a and c which you can solve.

For example, set . I get

Set . I get

I have to solve and . You can do this in various ways.

For instance, if I add the equations and , I get , so . Then plugging into , I get , so .

Thus,

28. Compute .

Set : This gives

Differentiate the last x-equation (using the Product Rule on the two terms on the right):

Set :

Differentiate the last x-equation:

Set :

Differentiate the last x-equation:

Solving (1) ( ) together with (3) ( ) gives and .

Solving (2) ( ) together with (4) ( ) gives and .

Thus, I have

The first integral is computed using ; the second and third use the inverse tangent formula:

29. Compute .

Set : This gives . Plug this into the last equation:

There are a number of ways to proceed. For instance, you can plug in two other values for x (say and ) to obtain two equations involving a and b, which you can solve simultaneously.

Alternatively, differentiate the last equation:

Set : This gives . Plug this into the last equation:

Set : This gives

Plugging , , and into the original decomposition, I have

30. How would you try to decompose using partial fractions?

31. What is wrong with the following "partial fractions decomposition"?

Partial fractions is the opposite of combining fractions over a
common denominator. In this case, the question is: "What
fractions would add up to ?" The
decompositions above *could* occur, since it has as the common denominator.

However, since you don't know beforehand what the fractions are, you must assume the "worst case" --- namely, that there might be an term. And in fact, there is --- if you work out the decomposition, it comes out to

Notice the term .

32. What is wrong with the following "partial fractions decomposition"?

The first two terms could be combined into a single term , so they're redundant. There is no reason to list the same denominator twice.

33. Find the partial fractions decomposition of

Try the decomposition

Clear denominators:

Set : I get . Plug it back in:

Differentiate:

Set : I get .

Differentiate again:

Set : I get .

Cancel the and -12 terms in the previous equation, then differentiate:

Set : I get . Since , it follows that .

Plug back in, then simplify the equation:

Set : I get . But , so .

Hence,

34. (a) Compute .

(b) Calvin Butterball tries to use the antiderivative from (a) to compute

He gets

Does this computation make sense? Why or why not?

(a) , so I try

Clear denominators:

Set : I get , or .

Set : I get , or .

Set : I get , or .

Therefore,

(b) The computation is incorrect, because the antiderivative is valid
only within intervals which don't contain the singularities at ,
, and . The interval
includes the singularity at . It is not legal to simply
"plug in the endpoints" --- to do this definite integral
correctly, you should set it up as two * improper
integrals*.

35. Find the area of the region under from to .

The area is

Here's the work for the antiderivative:

36. Compute .

Since is undefined at and is in the interval of integration (it's one of the endpoints), the integral is improper. I replace the "4" with a parameter a, then take the limit as a approaches 4 from the right.

37. Compute .

As , I have , and . Therefore, the integral diverges (to , since the term was negated).

38. Compute .

Here's the work for the two limits:

I used L'Hôpital's Rule to compute the second limit.

Here's the work for the antiderivative:

39. Compute .

is undefined. Hence, the integral diverges. (It doesn't diverge to or ; the limit is simply undefined.)

40. Compute .

Do the antiderivative:

Do the integral:

(Note that .)

The integral diverges to .

41. Compute .

Do the antiderivative:

Do the integral:

(Note that .)

The integral converges to .

42. Compute .

The first integral is

The second integral is

Therefore,

The graph of has a vertical asymptote at , but the (signed) area on each side is finite. The negative area to the left of partially cancels the positive area to the right of . Thus, the integral in this problem does not represent the actual area bounded by the graph, the x-axis, and the lines , , and .

In the next few problems, I'll use the following fact. If f and g are integrable functions on every finite interval and , then if converges, then converges.

Intuitively, if the bigger function's integral converges to a number, then the smaller function's integral must converge, because it's caught between that number and 0.

43. Prove that converges.

The interval of integration is . On this interval, , so , and . Therefore, if converges, then cconverges.

Now

Since converges, it follows that converges as well.

44. Prove that converges.

I have

(I built up from a known fact about trig functions to get an inequality with the function I'm trying to integrate on the "small" side.)

Now

Since converges and , it follows that converges as well.

45. (a) Show that the following integrals both diverge:

(It follows that diverges as well.)

(b) Show that
converges. (This is called the * Cauchy principal
value* of the integral; this problem shows that is not the same as .)

(a)

(b)

*To think is not enough; you must think of something.* -
*Jules Renard*

Copyright 2020 by Bruce Ikenaga