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7.1 - 7.1 Integration by Parts(Dated Let f(x and g(x be...

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7.1: Integration by Parts (Dated: October 12, 2011) Let f ( x ) and g ( x ) be differentiable functions. Then the product rule states that d dx £ f ( x ) g ( x ) / = f ( x ) g 0 ( x ) + g ( x ) f 0 ( x ). This shows that f ( x ) g 0 ( x ) + g ( x ) f 0 ( x ) has f ( x ) g ( x ) as an an- tiderivative, from which it follows that Z f ( x ) g 0 ( x ) dx + Z g ( x ) f 0 ( x ) dx = f ( x ) g ( x ). It is customary to write this formula as Z f ( x ) g 0 ( x ) dx = f ( x ) g ( x ) - Z g ( x ) f 0 ( x ) dx . It is called the formula for integration by parts. In terms of differentials, one has Z u dv = uv - Z v du , where u = f ( x ) and v = g ( x ). For definite integrals, Z b a f ( x ) g 0 ( x ) dx = £ f ( x ) g ( x ) / b a - Z b a g ( x ) f 0 ( x ) dx . Example 1 (Exercise 7.1.4 in the text) Let u = x and v 0 = e - x . Then u 0 = 1 and v = - e - x . Thus, by integration by parts, Z xe - x dx = - xe - x - Z - e - x dx = - xe - x - e - x + C . Example 2 (Exercise 7.1.7 in the text) Let u = x 2 and v 0 = sin π x. Then u 0 = 2 x and v = - cos π x π . Thus, by integration by parts, Z x 2 sin π x dx = - x 2 cos π x π - Z - 2 x cos π x π dx = - x 2 cos π x π + 2 π Z x cos π x dx . Let u 1 = x and v 0 1 = cos π x. Then u 0 1 = 1 and v 1 = sin π x π . Then, by integration by parts also, Z x cos π x dx = x sin π x π - Z sin π x π dx = x sin π x π + cos π x π 2 + C 1 . Consequently, Z x 2 sin π x dx = - x 2 cos π x π + 2 π x sin π x π + cos π x π 2 + C 1 = - x 2 cos π x π + 2 x sin π x π 2 + 2cos π x π 3 + C , where C = 2 C 1 π . Example 3 (Exercise 7.1.9 in the text) Let u = ln(2 x + 1) and v 0 = 1 . Then u 0 = 2 2 x + 1 and v = x . Thus, by integration by parts, Z ln(2 x + 1) dx = x ln(2 x + 1) - Z 2 x 2 x + 1 dx . Since Z 2 x 2 x + 1 dx = Z 2 x + 1 - 1 2 x + 1 dx = Z 2 x + 1 2 x + 1 dx - Z 1 2 x + 1 dx = Z 1 dx - 1 2 ln | 2 x + 1 |+ C = x - 1 2 ln | 2 x + 1 |+ C , it follows that Z ln(2 x + 1) dx = x ln(2 x + 1) - x + 1 2 ln | 2 x + 1 |- C .
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