Substitution calc II Hobart - Integration by Substitution...

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Integration by Substitution In this chapter we expand our methods of antidifferentiation. We have encoun- tered integrals which we have been unable to determine because we did not know an antiderivative for the integrand. The technique we discuss below is simply reversing the chain rule for derivatives. Remember, every derivative rule can be reversed to create an antidifferentiation rule. Our objectives are to • Use pattern recognition to find an indefinite integral. • Use a change of variables to find an indefinite integral. • Use a change of variables to evaluate a definite integral. All are techniques for integrating composite functions. The discussion is split into two parts: pattern recognition and change of variables. Both techniques in- volve a so-called u -substitution. With pattern recognition we perform the substi- tution mentally, and with change of variables we write the substitution in detailed steps. Pattern Recognition Recall that the chain rule for derivatives of composite functions states: THEOREM 3 . 1 (Chain Rule) . If y = F ( u ) and u = g ( x ) are differentiable then d dx [ F ( g ( x ))] = F 0 ( g ( x )) g 0 ( x ) or, equivalently, d dx [ F ( u )] = F 0 ( u ) du dx . Notice in this set-up that du dx = g 0 ( x ) ( 3 . 1 ) the corresponding differentials are 1 1 Though du dx is a derivative and a single entity, when working with differentials it is as if one can multiply each side of ( 3 . 1 ) by dx to obtain the equation in ( 3 . 2 ). du = g 0 ( x ) dx . ( 3 . 2 ) Reversing the chain rule, from the definition of an antiderivative, we get: THEOREM 3 . 2 (Substitution) . Assume F is an antiderivative of f (so F 0 = f ) and that g is differentiable. Then letting u = g ( x ) so du = g 0 ( x ) dx , we get Z f ( g ( x )) g 0 ( x ) dx = Z f ( u ) du = F ( u ) + C = F ( g ( x )) + c . ( 3 . 3 )
math 131 , days 9 and 10 u - substitution 2 Once we know F is and antiderivative of f , then by definition of antiderivative we have R f ( u ) du = F ( u ) + c . . . for example, R cos u du = sin u + c no matter what the differentiable function u is . This is the whole point of substitution whether it is done mentally or written out: Convert the problem into one that we can solve ‘just by looking at it.’ To apply Theorem 3 . 2 directly, we must recognize the presence of f ( g ( x )) and g 0 ( x ) , i.e., f ( u ) and du dx . Note that the composite function in the integrand has an outside function f and an inside function g . Moreover, the derivative g 0 ( x ) must also be present as a factor of the integrand so that ( 3 . 3 ) becomes: Z outter z}|{ f ( g ( x ) | {z } inner ) inner deriv z }| { g 0 ( x ) dx . EXAMPLE 3 . 1 (Recognizing the pattern) . Determine Z cos ( x 3 ) 3 x 2 dx . SOLUTION. There is a function which is literally ‘inside’ the parentheses in this prob- lem. So the inside function is u = x 3 and its derivative, 3 x 2 is a factor in the inte- grand. Z cos ( u z}|{ x 3 ) du z }| { 3 x 2 dx . In other words, 2 the integral can be thought of as 2 Let u = x 3 and du = 3 x 2 dx .

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