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Published byKyra Hannah Modified over 2 years ago

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Lesson 7-1 Integration by Parts

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Derived from the Product Rule in Differentiation D(uv) = v × du + u × dv ∫ d(uv) = ∫v du + ∫u dv uv = ∫v du + ∫u dv ∫u dv = uv – ∫v du Used to make integrals simpler

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Integration by Parts Strategies Select u so that taking its derivative makes a simpler function Let dv be something that can be integrated 1)Use derivative to drive a polynomial function to zero 2)Reduce polynomials to get a u-substitution 3)Use derivative to get the original integral and the simplify using addition/subtraction

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Walk through Example Solve: x cos x dx If we let u = x and dv = cos x dx, then du = dx and v = sin x ∫u dv = uv – ∫v du x cos x dx = x sin x - sin x dx x cos x dx = x sin x + cos x + C

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7-1 Example 1 ∫ x e x dx let u = x and dv = e x dx du = dx and v = e x = x e x – ∫ e x dx = = x e x – e x + C = e x (x + 1) + C

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7-1 Example 2 ∫ x ln x dx let u = ln x and dv = x dx du = dx/x and v = ½ x² = ½x² ln x - ∫ ½ x ² dx/x = ½x² ln x - ¼x² + C = ¼x² (2ln x – 1) + C = ½x² ln x - ∫ ½ x dx

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7-1 Example 3 ∫ x sin 3x dx let u = x and dv = sin 3x dx du = dx and v = -⅓cos 3x = -⅓x cos 3x - ∫ -⅓ cos 3x dx = -⅓x cos 3x +⅓(⅓ sin 3x) + C = (1/9) (sin 3x – 3x cos 3x) + C = -⅓x cos 3x +⅓ ∫ cos 3x dx

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Summary & Homework Summary: –Integration by parts allows us to solve some previously unsolvable integrals –Methods: Use derivative to drive a polynomial function to zero Reduce polynomials to get a u-substitution Use derivative to get the original integral and the simplify using addition/subtraction Homework: –pg 480 – 482: Day 1: 3, 4, 7, 9, 36;

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Integration by Parts – Repeated Use Sometimes we have to use the method of integration by parts several times to get an integral that we can solve or to get it to repeat Using a table to record our differentiations and integrations can help keep things straight

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7-1 Example 4 ∫ sin x e x dx let u = sin x and dv = e x dx du = cos x dx and v = e x = sin x e x – ∫ cos x e x dx = let u = cos x and dv = e x dx du = - sin x dx and v = e x = sin x e x – (cos x e x - ∫ - sin x e x dx ) = 2 ∫ sin x e x dx = sin x e x – cos x e x = ½ ( sin x e x – cos x e x ) + c Remember to keep the () in the problem!

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7-1 Example 5 ∫ cos 2x e x dx let u = cos 2x and dv = e x dx du = -2sin 2x dx and v = e x = cos 2x e x – ∫ -2sin 2x e x dx = let u = sin 2x and dv = e x dx du = 2cos 2x dx and v = e x = cos 2x e x + 2 (sin 2x e x - ∫ 2cos 2x e x dx ) 5 ∫ cos 2x e x dx = cos 2x e x + 2sin 2x e x + C = ( e x /5) ( cos 2x + 2sin 2x ) + C Remember to keep the () in the problem! ∫ cos 2x e x dx = cos 2x e x + 2sin 2x e x - 4 ∫ cos 2x e x dx

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7-1 Example 6 ∫ x² sin x dx let u = x² and dv = sin x dx du = 2x dx and v = -cos x = -x² cos x – ∫ -cos x 2x dx = let u = x and dv = cos x dx du = dx and v = sin x = -x² cos x + 2(x sin x - ∫ sin x dx ) ∫ x² sin x dx = -x² cos x + 2(x sin x – (-cos x)) + C = -x² cos x + 2x sin x + 2cos x + C = (2 - x²) cos x + 2x sin x + C Remember to keep the () in the problem!

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7-1 Example 7 Find the integral of ∫ (6x³ + 3x² - 5x – 7) e x dx DifInt 6x³ + 3x² - 5x - 7+exex

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7-1 Example 7 Find the integral of ∫ (6x³ + 3x² - 5x – 7) e x dx DifInt 6x³ + 3x² - 5x - 7+exex 18x² + 6x - 5-exex 36x + 6+exex 36-exex 0+exex ∫ (6x³ + 3x² - 5x – 7) e x dx = = e x (6x³ + 3x² - 5x - 7) – e x (18x² + 6x – 5) + e x (36x + 6) – e x (36) + C = e x (6x³ - 15x² + 25x - 28) + C

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Summary & Homework Summary: –Integration by parts allows us to solve some previously unsolvable integrals –Methods: Use derivative to drive a polynomial function to zero Reduce polynomials to get a u-substitution Use derivative to get the original integral and the simplify using addition/subtraction Homework: –pg 480 – 482: Day 1: 3, 4, 7, 9, 36; Day 2: 1, 14, 19, 51

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