MATH 116 ACTIVITY 4:Integration by parts -practice, an extension, and a special case

WHY:The integration by parts technique introduced in the last class extends our antiderivative methods to handle a lot more situations. It is also complicated enough that you don't learn it in one day - and provides a clear problem-solving situation (you have to decide whether to use parts, decide what parts to use, and then use some still-complicated tools to carry out the work). In addition the two-or-more-step case and the "hidden split" case should be seen before we move on.

LEARNING OBJECTIVES:

1.Be able to recognize an integration by parts situation.

2. Be able to use integration by parts-basic and multi-step

3.Work as a team, using the team roles.

CRITERIA:

1.Success in completing the exercises.

2.Success in working as a team and in filling the team roles.

RESOURCES:

1.Your text - especially section 8.2

3.Class notes and examples

4The team role desk markers (handed out in class for use during the semester)

5.40 minutes

PLAN:

1.Select roles, if you have not already done so, and decide how you will carry out steps 2 through 4

2. Read through the discussion and models given here

3.Complete the exercises given here - be sure all members of the team understand and agree with all the results in the recorder's report.

3.Assess the team's work and roles performances and prepare the Reflector's and Recorder's reports including team grade .

DISCUSSION:

Integration by parts is a useful tool for calculating indefinite integrals (and so, of course, for calculating definite integrals) that do not directly match our basic formulas. It is based on the product rule for derivatives, and is, in effect, a product rule for antiderivatives.
Usually the indication that we want to use integration by parts is that when we try to match a given problem to one of our basic formulas, there is extra variable material in the problem (beyond that needed for the u'), so that we would have a possible antiderivative except for extra variable pieces. We wind up splitting the problem into a v' part (which we can integrate) and a u part (which we wish would go away). The v' is, typically, "the worst thing there that we can integrate" (exponentials, trig functions, radicals - all with their u' parts - once we are already in the "parts" situation) and the u is "what's left".
The formula is then When it works, the new integral (on the right) is easier than the original integral (on the left.) (If it's worse, that usually signals that the wrong choices were made in identifying u and v')
The extension Sometimes the new integral is not one that we can directly calculate, but is closer (the pesky old u didn't go away completely, but is reduced), and a second use of the parts technique will give us an integral that we can handle - or maybe a couple more steps - see model 2 below.
The special case Sometimes our algebraic notation has part of the v' hidden in the u and we have to separate them (writing x2 as x x or x5 as x3 x2 is the typical sort of thing) because the u' of the formula used to get v involves part of the variable mess that required parts in the first place (see model 3 below)

MODELS

1. If we try to use the formula, we have u = 3x, u' = 3 - but there's an extra x. (If we try to use any formula with x = u, we have an extra cos(3x)) so this is a candidate for parts. We'd try u = x and v' = cos(3x) or
u = cos(3x) and v'=x. Since the integral of cosine is sine (no messier), we often use trig functions for v' (integrating cos(3x) is less basic than integrating x, so the "messiest thing we can integrate" rule also suggests v'=cos(3x)) so we try that:
u = x, v' = cos(3x) so
u' = 1 v = This gives us and we can calculate the last integral using the rule (with u = 3x, u'=3), giving so we get

Notice that if we had we would not use the integration by parts techniques at all, because we would use the formula with u = 3x2 and u' = 6x - the x is used by the basic formula and is not extra.

2. . If we try to use the formula with u = -x, we have u' = -1 and there is an extra x2 in the integral - so we try parts. Since there is an exponential, we try
u = v' =
u’ = 2x v = [we hold off on the +C] and we get step 1
Step 2 The new integral still has an extra x in it (derivative of -x is -1) - but that is reduced from an x2 in the original problem, so integration by parts gave us a simpler integral. If we try again with this integral, maybe things will get better.
Looking only at the new integral (we'll plug it back into the problem after). we try integration by parts with
u = x, v' =
u' = 1, v = (same as before - because the v' is the same) and our new integral is found as (Again, leaving off the +C for the moment) - substituting in our step 1 =

3. This looks like with u = - but then u' = 2x and we have an extra x2 - in the x3 . We should again try integration by parts, but the u is not obvious in the original problem - it's hidden as part of the x3 . Using the "v' as the worst thing we can integrate" we try
- because we can't integrate without the x (the u' for our un)
u' = 2x (saving the +C for later) and we get

Some remarks on identifying u, v’:We select u, v’ so that will be simpler than , Since exponential and trig functions have antiderivatives that are no messier than the functions, they are candidates for v’. Since the logarithm is hard to deal with, ln(x) (or powers of ln(x)) is always a prime candidate for u. Powers of x are usually candiates for u.

EXERCISES:

Evaluate each of the following indefinite integrals. The work may use standard antiderivative rules, integration by parts, multi-step integration by parts, or "hidden u" integration by parts.

1.

2.

3.

4.

CRITICAL THINKING QUESTIONS:(answer individually in your journal)

1.Which of the added pieces (extension to two or more steps, "hidden u") of the integration by parts technique do you find harder to deal with? Why?

2. We now have eight antiderivative rules (the basic 5, constant coefficient and addition/subtraction, and now integration by parts). Which are easiest to recall and use? Which are hardest? Why?

3.Did discussion wit your teammates help to clear up questions on integration by parts?

SKILL EXERCISES

Text p. 454 # 15-16, 19-21, 24 – 28, 40-41, 43