Section 7.1: Integration by Parts

Transcription

1 Section 7.1: Integration by Parts 1. Introuction to Integration Techniques Unlike ifferentiation where there are a large number of rules which allow you (in principle) to ifferentiate any function, the operation of integration is much more technical an there are in fact many functions sin () for which there just on t eist algebraic antierivatives (take for eample, the TI-89 just gives you back the same function suggesting that there is in fact no algebraic antierivative for this function). There are two ways we are going to try to further our unerstanings an abilities in integration. (i) Derive rules an techniques, like we i with erivatives, to allow us to integrate as many ifferent types of functions as possible. (ii) Look at ifferent ways to write functions own for which it may be easier to fin integrals (linear approimations etc). The majority of Chapter 7 focuses on eriving new rules an techniques for integration (Chapter 11 will focus on the problem of realizing functions in new ways). We start by briefly recalling substitution which arose as an attempt to reverse the chain rule. 2. Review of Integration by Substitution Result 2.1. If u g() is a ifferentiable function whose range is an interval I an f is continuous on I, then f(g())g () f(u)u. Recall that the basic approach to problems in integration by substitution are as follows: (i) Ientify an insie function g() an an outsie function f() so that the integran looks something like f(g())g () (ii) Set u g() an then or u g (), 1 g () u. In the original integran, replace g() by u an by 1 u. g () (iii) Simplify the equation. Provie no mistakes have been mae, you shoul have an integran solely in terms of u. If not, either a mistake has been mae or the substitution you have use oes not work or there may be further simplifications which can be use to eliminate from the equation. (iv) Evaluate the integral as an integral with respect to u. 1

2 2 (v) Substitute back in for. We illustrate the process with a couple of eamples: Eample 2.2. Evaluate the following integrals. (i) 1+4 ( ) Choose u , then u/ 1 + 4, or u/(1 + 4). Making the substitution, we get ( ) 1 u u 1 u u u 1/2 u 2u 1/2 + C 2 ( ) + C. (ii) + 1 Choose u 1 +, then u/ 1 or u. Making the substitution, we get uu which looks like something we still cannot integrate. However, u + 1, so u 1 giving, uu (u 1)u 1/2 u u 3/2 u 1/2 u 2 u/2 3 2 u3/2 + C 2 ( + 1)/2 3 2 ( + 1)3/2 + C. Question 2.3. The iea of substitution came from trying to reverse the chain rule. Are there any other rules we can try to reverse? 3. Integration by Parts Recall, the prouct rule says the following: if f() an g() are ifferentiable functions, then the erivative of f()g() is f ()g()+f()g () i.e f()g() ( f())g() + f()( g()). This means that f()g() ( f())g() f()( g()). If we integrate both sies an apply the funamental theorem of calculus, we get, (f()g())) ( f())g()) (f()( g())).

3 But, (f()g()) ( f())g() f()g() ( f())g() (f()( g())). This means in orer to calculate the integral of a function f() multiplie by a erivative of a function g (), we can apply this formula i.e f()g () f()g() f ()g(). This is not quite a prouct rule, but it can be very hany if we have a prouct of functions, one of which has an easy integral an the other an easy erivative. We shall look at the process of integration by parts in etail. (i) In the integran, look for a prouct of functions. Set u f(), a function whose erivative is easier than f(), an set v g (), a function for which it is easy to fin an antierivative. If in oubt, use LIPET (logarithm, Inverse Trip, Polynomial, Eponential, Trig Function) in your choice for f(). (ii) Fin v g () an u f (). (iii) To calculate f()g () we use the formula f()g () f()g() f ()g(). in terms of u an v, we have u v uv v u. The integral on the right han sie in principle shoul be easier than the integral on the left han sie, so we integrate. We illustrate the process with a number of eamples. Eample 3.1. Use integration by parts to evaluate the following integrals. (i) e Here the function will become easier if ifferentiate, so we set u an v/ e. Then, u/ 1 an v e. Applying the formula, we get, e e e e e + C. (ii) cos () Here the function will become much easier if ifferentiate, so we set u an v/ cos (). Then u/ 1 an v (1/) sin(). Applying integration by parts, we get cos () 1 sin () sin () 3

4 4 sin () + 1 cos () + C. 2 (iii) ln () Here it oes not look like the prouct of two functions. However, notice that any function f() is equal to 1 f(). In this case, we can eliminate ln () through ifferentiation, so we set u ln () an v/ 1 giving, u/ 1/ an v. Then applying the formula, we get ln() ln () 1 ln () + C. (iv) sin 1 (). This question is similar to the previous one. We set u sin 1 () an v/ 1 giving, u/ 1/ 1 2 an v. Then applying the formula, we get sin 1 () sin 1 () 1 2. To solve this last integral, we nee to use substitution. We take u 1 2, then u/ 2 or u/( 2) giving u 2 u 1 u 1/2 u u 1/ Putting these together, we get sin 1 () sin 1 () C. (v) 4 (ln()) 2 Here we cannot integrate (ln ()) 2, so we shall set u (ln ()) 2 an v/ 4. Then u/ 2 ln()/ an v /. Applying the formula, we get: 2 ln() 4 (ln()) 2 [(ln ()) 2 [(ln ()) 2 2 ln () 4 This secon integral is also not possible without integration by parts, so again we set v/ 4 an u ln (), so v / an u/ 1/. Applying the formula, we get: [ 4 ln() ln () [ ln () 1 [ 4 ln () 2

5 Putting these two together, we get 4 (ln()) 2 [(ln ()) 2 2 [(ln()) 2 2 [(ln()) 2 [ ln() 2 ln () Reuction Formulas ln () 4 The last eample we consiere require performing integration by parts twice. In general, when a function involves large powers, in practice, integration by parts can be use to erive an antierivative. However, in many cases it is not practical. What is practical however is fining instea a formula which one can use a number of times rather than following the same process continually. Fining a formula using integration by parts which reuces the compleity of an integral without actually solving it is calle fining a reuction formula. We illustrate by eample. Eample 4.1. Show that (ln ()) n (ln ()) n n (ln ()) n 1. Why oes this help? Even though we cannot fully evaluate it, we shall use integration by parts. We choose u (ln ()) n an v/ 1 giving u/ n(ln ()) n 1 / an v. Applying the integration by parts formula, we get n(ln ()) (ln ()) n (ln ()) n n 1 (ln ()) n n (ln ()) n 1. This formula is helpful because it allows us to evaluate such integrals without having to apply integration by parts continually. For eample, if n 3, then (ln ()) 3 (ln ()) 3 3 [ (ln ()) 3 3 (ln ()) 2 2 (ln ()) 2 (ln ()) 1 [ (ln ()) 3 3 (ln ()) 2 2( ln() ).

6 6. Definite Integrals As with substitution, integration by parts can be use to evaluate efinite integrals in eactly he same way using the funamental theorem of calculus. b a f()g () We illustrate with an eample. [ f()g() b f()g () a Eample.1. Evaluate the efinite integral 1 0. In this case, we take u an v/. Then we have, u/ 1 an v / ln(). Applying integration by parts, we have 1 0 [ ln () [ ln () (ln()) ln () 1 ln() 4 ln () 2 0