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Unformatted text preview: (3/20/08) Section 6.5 Integrals of powers and indefinite integrals Overview: In this section we use Part I of the Fundamental Theorem of Calculus to derive a formula for integrals of powers y = x n with n negationslash = 1 . Then we introduce the term “indefinite integral” for “antiderivative.” Topics: • Integrals of y = x n with n negationslash = 1 • Indefinite integrals Integrals of y = x n with n negationslash = 1 We can evaluate a definite integral of a function y = f ( x ) if we can find a function y = F ( x ) whose derivative is f . Such a function F is called an antiderivative of f . In particular, if F is continuous and f prime is piecewise continuous on an interval and F prime ( x ) = f ( x ) in the interior of the interval, then by Part I of the Fundamental Theorem (Theorem 2 of Section 6.3), integraldisplay b a f ( x ) dx = integraldisplay b a F prime ( x ) dx = F ( b ) F ( a ) (1) for any numbers a and b in the interval. We use this approach to derive the following theorem. Theorem 1 For any constant n negationslash = 1 and for a and b in an interval where x n is defined, integraldisplay b a x n dx = bracketleftBig 1 n + 1 x n +1 bracketrightBig b a = 1 n + 1 b n +1 1 n + 1 a n +1 . (2) In formula (2) we used the symbol bracketleftBig F ( x ) bracketrightBig b a for F ( b ) F ( a ). Notice that (2) reads integraldisplay b a 1 dx = b a for n = 0 and reads integraldisplay b a x dx = 1 2 b 2 1 2 a 2 for n = 1. Theorem 1 does not apply if n = 1. We will see in Section 6.6 that integrals of x 1 are given by logarithms. Proof of Theorem 1: Let n be any constant negationslash = 1. To apply (1) with f ( x ) = x n , we need a function F whose derivative is x n . Notice that in finding the derivative d dx ( x n ) = nx n 1 , we first multiply x n by the exponent n and then decrease the exponent by 1. Accordingly, we should be able to find a function whose derivative is x n by performing the opposite steps in the opposite order. We raise the exponent in x n by 1 to n + 1 and then divide by the new exponent to obtain F ( x ) = 1 n + 1 x n +1 . This is, in fact, an antiderivative of x n because F prime ( x ) = d dx parenleftbigg 1 n + 1 x n +1 parenrightbigg = 1 n + 1 [( n + 1) x n ] = x n . (3) Setting f ( x ) = x n and F ( x ) = 1 n + 1 x n +1 in (1) gives (2) . QED 218 Section 6.5, Integrals of powers and indefinite integrals p. 219 (3/20/08) Example 1 What is the value of the integral integraldisplay 1 1 x 2 dx ? Solution By (2) with a = 1 ,b 2, and n = 2, for which n + 1 = 3, integraldisplay 1 1 x 2 dx = bracketleftBig 1 3 x 3 bracketrightBig 1 1 = 1 3 (1 3 ) 1 3 ( 1) 3 = 2 3 . square We can apply Theorem 1 to linear combinations y = Ax n + Bx m of powers of x , as in the next example, by writing for n negationslash = 1 and m negationslash = 1, integraldisplay b a ( Ax n + Bx m ) dx = bracketleftbigg A n + 1 x n +1 + B m + 1 x m +1 bracketrightbigg b a . (4) Example 2 Evaluate integraldisplay 2 1 (4 x 1 / 3 + 6...
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 Summer '08
 Eggers
 Math, Derivative, Fundamental Theorem Of Calculus, Integrals, 1 m, 0 k

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