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Math Help - Integration Problem

  1. #1
    Member Em Yeu Anh's Avatar
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    Question Integration Problem

    There are two parts to this question. First, find f(4) if \int_0^{x^2}f(t)dt = x\cos{{\pi}x}

    The next part asks to Find f({\pi}/2) if f satisfies the conditions (i) f is positive and continuous, and (ii) the area under the curve y = f(x) from x = 0 to x = a is:
    \frac{a^2}{2} + \frac{a}{2}sina+\frac{\pi}{2}cosa

    I don't think I'm doing this correctly. Same error for both parts, I tried applying the fundamental thereom but end up having to deal with f(0) which is unknown.
    For part one I so far have:
    \int_0^{x^2}f(t)dt = x\cos{{\pi}x} = f(x^2) - f(0)

    2cos2{\pi} = f(4) - f(0)

    2 + f(0) = f(4) ??

    Anyone mind guiding me in the correct direction? Thanks! Know it's a lot to read.
    Last edited by Em Yeu Anh; November 16th 2009 at 08:24 PM.
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  2. #2
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    Quote Originally Posted by Em Yeu Anh View Post
    There are two parts to this question. First, find f(4) if \int_0^{x^2}f(t)dt = x\cos{{\pi}x}

    [snip]
    For part one I so far have:
    \int_0^{x^2}f(t)dt = x\cos{{\pi}x} = f(x^2) - f(0)

    2cos2{\pi} = f(4) - f(0)

    2 + f(0) = f(4) ??

    [snip]
    I have time to do the first. You are on the wrong track. You need to differentiate both sides with respect to x (for the left hand side, use the chain rule and the Fundamental Theorem of Calculus). This will give you f(x).
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  3. #3
    Member Em Yeu Anh's Avatar
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    Thank you,
    alright I'll try that again:
    \frac{d}{dx}\int_0^{x^2}f(t)dt = -{\pi}xsin{\pi}x + cos{\pi}x

    f(x^2)*2x = -{\pi}xsin{\pi}x + cos{\pi}x

     f(x^2) = \frac{-{\pi}sin{\pi}x}{2} + \frac{cos{\pi}x}{2x}

    f(4) = \frac{1}{4}

    Assuming the same method will be applicable to part 2,

     \frac{d}{dx}\int_0^af(x)dx = a + \frac{acosa}{2} - \frac{{\pi}sina}{2}

    f({\pi}/2) = \frac{\pi}{2} + \frac{({\pi}/2)cos({\pi}/2)}{2} - \frac{{\pi}sin({\pi}/2)}{2}

    f({\pi}/2) = 0
    Last edited by Em Yeu Anh; November 16th 2009 at 08:31 PM. Reason: latex errors
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  4. #4
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    Quote Originally Posted by Em Yeu Anh View Post
    Thank you,
    alright I'll try that again:
    \frac{d}{dx}\int_0^{x^2}f(t)dt = -{\pi}xsin{\pi}x + cos{\pi}x

    f(x^2)*2x = -{\pi}xsin{\pi}x + cos{\pi}x

     f(x^2) = \frac{-{\pi}sin{\pi}x}{2} + \frac{cos{\pi}x}{2x}

    f(4) = \frac{1}{4}

    Assuming the same method will be applicable to part 2,

     \frac{d}{d{\color{red}a}}\int_0^af(x)dx = a + \frac{acosa}{2} - \frac{{\pi}sina}{2} Mr F says: 1. Note the red a, and 2. Your derivative of the right hand side with respect to a is missing a term.

    f({\pi}/2) = \frac{\pi}{2} + \frac{({\pi}/2)cos({\pi}/2)}{2} - \frac{{\pi}sin({\pi}/2)}{2}

    f({\pi}/2) = 0
    ..
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  5. #5
    Member Em Yeu Anh's Avatar
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    Not sure what I missed; Does \frac{a}{2}sin(a) require a product rule?
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  6. #6
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    Quote Originally Posted by Em Yeu Anh View Post
    Not sure what I missed; Does \frac{a}{2}sin(a) require a product rule?
    Well, a is the variable you're differentiating with respect to. And what you're differentiating is a product of two functions of a. So what do you think ....?
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  7. #7
    Member Em Yeu Anh's Avatar
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    Of course, another problem from my error of thinking that it was differentiated with respect to x, I misinterpreted "a" as just being a constant.

    Redo once again:
    <br />
\frac{d}{da}\int_0^af(x)dx = a + \frac{a}{2}cos(a) + \frac{sin(a)}{2} - \frac{\pi}{2}sin(a)

    f(\frac{\pi}{2}) = \frac{1}{2}

    Hopefully no more mistakes this time.
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  8. #8
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    Quote Originally Posted by Em Yeu Anh View Post
    Of course, another problem from my error of thinking that it was differentiated with respect to x, I misinterpreted "a" as just being a constant.

    Redo once again:
    <br />
\frac{d}{da}\int_0^af(x)dx = a + \frac{a}{2}cos(a) + \frac{sin(a)}{2} - \frac{\pi}{2}sin(a)

    f(\frac{\pi}{2}) = \frac{1}{2}

    Hopefully no more mistakes this time.
    That looks OK.
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  9. #9
    Member Em Yeu Anh's Avatar
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    Excellent! Thanks, I wouldn't have had a chance at getting this correct without your help.
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