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Thread: Integrating the second derivative

  1. #1
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    Integrating the second derivative

    given f''(x) = -25sin(5x) and f'(0) = 3 and f(0) = -2, find f(pi/6)

    I am trying to figure out how to integrate the second derivative into the first derivative and make it 3, then integrate into a function of 0 that equals -2.

    How do I integrate the second derivative into the first derivative?
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  2. #2
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    Quote Originally Posted by derekjonathon View Post
    given f''(x) = -25sin(5x) and f'(0) = 3 and f(0) = -2, find f(pi/6)

    I am trying to figure out how to integrate the second derivative into the first derivative and make it 3, then integrate into a function of 0 that equals -2.

    How do I integrate the second derivative into the first derivative?
    $\displaystyle f''(x) = -25\sin{(5x)}$

    $\displaystyle f'(x) = \int{-25\sin{(5x)}\,dx}$

    $\displaystyle f'(x) = 5\cos{(5x)} + C$.


    You know that $\displaystyle f'(0) = 3$, so

    $\displaystyle 3 = 5\cos{(5\cdot 0)} + C$

    $\displaystyle 3 = 5 + C$

    $\displaystyle C = -2$.


    So $\displaystyle f'(x) = 5\cos{(5x)} - 2$

    $\displaystyle f(x) = \int{5\cos{(5x)} - 2\,dx}$

    $\displaystyle f(x) = \sin{(5x)} - 2x + C$.


    You know that $\displaystyle f(0) = -2$, so

    $\displaystyle -2 = \sin{(5\cdot 0)} - 2\cdot 0 + C$

    $\displaystyle C = -2$.


    Therefore

    $\displaystyle f(x) = \sin{(5x)} - 2x - 2$.


    Now find $\displaystyle f\left(\frac{\pi}{6}\right)$.
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  3. #3
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    Thanks!

    It seems easy when you explain it that way!
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