note10 - Math021, week 10 Integration by Substitution...

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Math021, week 10 Integration by Substitution Theorem 10.1 If f and g are functions, f ( g ( x )) is an antiderivative of f 0 ( g ( x )) g 0 ( x ) . That is, Z f 0 ( g ( x )) g 0 ( x ) dx = f ( g ( x )) + C. proof: nothing different from the orinary chain rule. Example 10.2 Evaluate Z 2 x (1 + x 2 ) 8 dx. solution: Let g ( x ) = 1 + x 2 and h ( x ) = x 8 . We are asking for an antiderivative of g 0 ( x ) h ( g ( x )). Now if f ( x ) = 1 9 x 9 , f is an antiderivative of h . In other words, we are asking for an antiderivative of g 0 ( x ) f 0 ( g ( x )). According to the version of chain rule above, an antiderivative of it is f ( g ( x )) = 1 9 (1 + x 2 ) 8 . Thus, Z 2 x (1 + x 2 ) 8 dx = 1 9 (1 + x 2 ) 9 + C. Remark 10.3 The previous procedure can easily be memorized by putting u = x 2 + 1 , du = 2 xdx . Then, Z 2 x (1 + x 2 ) 8 dx = Z u 8 du = 1 9 u 9 + C = 1 9 (1 + x 2 ) 9 + C. Example 10.4 Evaluate Z e 4 x dx. Let u = 4 x so that du = 4 dx . Then, R e 4 x dx = R e u 1 4 du = 1 4 e u + C = 1 4 e 4 x + C. 1
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Example 10.5 Evaluate Z e sin x cos xdx. solution: Let u = sin x so that du = cos xdx . Then, R e sin x cos xdx = R e u du = e u + C = e sin x + C. Example 10.6 Evaluate the area of the region bounded by the graph of f ( x ) = e sin x cos x , the x-axis, the lines x = 0 and x = π/ 2 . solution: Note that f ( x ) 0 for 0 x π/ 2. Hence, the area of the given region is Z π/ 2 0 e sin x cos xdx. We saw that g ( x ) = e sin x is an antiderivative of f ( x ) = e sin x cos x . By the fundamental theorem of calculus, Z π/ 2 0 e sin x cos xdx = e sin x | π/ 2 0 = e 1 - e 0 = e - 1 which is the area of the given region. Remark 10.7
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note10 - Math021, week 10 Integration by Substitution...

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