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Denebel

  • 2 years ago

evaluate the integral analytically.

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  1. Denebel
    • 2 years ago
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    |dw:1327548906549:dw|

  2. shinigami1m
    • 2 years ago
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    use integration by parts

  3. Zarkon
    • 2 years ago
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    the integrand is positive on 0 to pi/2 so the answer has to be a positive number

  4. Denebel
    • 2 years ago
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    The answer is (pi^2/8) - (1/2)

  5. Denebel
    • 2 years ago
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    according to the back of the book

  6. Zarkon
    • 2 years ago
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    your book is correct ;)

  7. Denebel
    • 2 years ago
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    Ok, how do I get there?

  8. Zarkon
    • 2 years ago
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    integration by parts ;)

  9. ash2326
    • 2 years ago
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    sorry it'd be -x^2 cos x + \(\int\limits_{0}^{\pi/2} 2x \cos xdx\) -x^2 cos x+2x sin x-\(\int\ limits_{0}^{pi/2} 2 sin x dx so we get -x^2 cos x+ 2x sin x+2 cos x 2x sin x- cos x(x^2 -2) now insert the limits [ 2* pi/2 * 1-0]-[0-1*(0-2)] pi -2

  10. Zarkon
    • 2 years ago
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    that is correct if it was \[x^2\sin(x)\] but it is \[x^2\sin(2x)\]

  11. ash2326
    • 2 years ago
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    sorry i thought it's x^2 sin x let me put the correct one -x^2 cos 2x /2+ ∫ 0 π/2 xcos2xdx -x^2 cos 2x /2 + x sin 2x/2 -∫ 0 π/2 sin2x/2dx -x^2 cos 2x/2 + x sin 2x /2 +cos 2x /4 now inserting the limits [pi^2/8+0-1/4]-[1/4] pi^2/8-1/2 π^2 /8-1/2

  12. Denebel
    • 2 years ago
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    Thanks!!

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