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Math Help - Help with Antiderivatives

  1. #1
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    Help with Antiderivatives

    find f(x) if f''(x)=2+cos x, f'(0)=-1, f(\pi/2)=0

    This is what I get:
    f'(x)=2x+sin x -1
    f(x)=x^2-cos x-x+(\frac{\pi}{2})-(\frac{\pi}{2})^2

    The correct answer is: x^2-cos x-\frac{1}{2}\pi x

    What did I do incorrectly?
    Last edited by yoman360; September 26th 2009 at 09:10 AM.
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  2. #2
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    Quote Originally Posted by yoman360 View Post
    find f(x) f''(x)=2+cos x, f(0)=-1, f(\pi/2)=0

    This is what I get:
    f'(x)=2x+sin x -1
    f(x)=x^2-cos x-x+(\frac{\pi}{2})-(\frac{\pi}{2})^2

    The correct answer is: x^2-cos x-\frac{1}{2}\pi x

    What did I do incorrectly?
    I don't know how you got f'(x) = 2x + \sin{x} - 1, because your two initial conditions are for f, not f'.


    You should have

    f''(x) = 2 + \cos{x}

    f'(x) = 2x + \sin{x} + C_1

    f(x) = x^2 - \cos{x} + C_1x + C_2.


    Now apply the initial conditions and you'll get two equations to solve simultaneously for C_1 and C_2.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    I don't know how you got f'(x) = 2x + \sin{x} - 1, because your two initial conditions are for f, not f'.


    You should have

    f''(x) = 2 + \cos{x}

    f'(x) = 2x + \sin{x} + C_1

    f(x) = x^2 - \cos{x} + C_1x + C_2.


    Now apply the initial conditions and you'll get two equations to solve simultaneously for C_1 and C_2.
    oops i made a typo it is f'(0)= -1
    NOT f(0)=-1
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  4. #4
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    i still need help
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  5. #5
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    Quote Originally Posted by yoman360 View Post
    find f(x) if f''(x)=2+cos x, f'(0)=-1, f(\pi/2)=0

    This is what I get:
    f'(x)=2x+sin x -1
    f(x)=x^2-cos x-x+(\frac{\pi}{2})-(\frac{\pi}{2})^2

    The correct answer is: x^2-cos x-\frac{1}{2}\pi x

    What did I do incorrectly?
    based on the equation for f''(x) and the given initial conditions, the "correct" answer you posted is incorrect.

    your solution for f(x) is correct.
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  6. #6
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    Quote Originally Posted by skeeter View Post
    based on the equation for f''(x) and the given initial conditions, the "correct" answer you posted is incorrect.

    your solution for f(x) is correct.
    ok thanks i tried this problem numerous times and i keep getting my answer which is the correct answer i guess.
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