7.1 - 7.1: Integration by Parts (Dated: October 12, 2011)...

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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)
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7.1 - 7.1: Integration by Parts (Dated: October 12, 2011)...

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