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Math Help - derivative/integral problem.

  1. #1
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    derivative/integral problem.

    F(x)+5=\int_{2}^{x}sin(\frac{\pi t}{4})dt
    what is F'(2)+F(2)


    -my steps
    F'(x)=sin(\frac{\pi 2}{4}) - 5=sin(\frac{\pi }{2})-5=1-5=-4
    then i realized something...
    since a derivative and an integral are inverses. Isn't F'(x)+F(x) always gonna equal 0?

    I am right or going crazy?

    All thanks is appreciated.
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  2. #2
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    F'(x) = \sin\left(\frac{\pi x}{4}\right)

    F'(2) = 1

    F(2) + 5 = \int_2^2 \sin\left(\frac{\pi t}{4}\right) \, dt

    F(2) + 5 = 0

    F(2) = -5

    so ...

    F'(2) + F(2) = -4


    F'(x) and F(x) are not inverses, and their sum is not always 0.


    ... you're crazy?
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by dandaman View Post
    F(x)+5=\int_{2}^{x}sin(\frac{\pi t}{4})dt
    what is F'(2)+F(2)


    -my steps
    F'(x)=sin(\frac{\pi 2}{4}) - 5=sin(\frac{\pi }{2})-5=1-5=-4
    then i realized something...
    since a derivative and an integral are inverses. Isn't F'(x)+F(x) always gonna equal 0?

    I am right or going crazy?

    All thanks is appreciated.
    You're not quite there.

    Recall by the FTC, \frac{\,d}{\,dx}\left[\int_a^xf\!\left(t\right)\,dt\right]=f\!\left(x\right)

    So in the case of F\!\left(x\right)+5=\int_2^x\sin\!\left(\tfrac{\pi  }{4}t\right)\,dt, we see that F'\left(x\right)=\sin\!\left(\tfrac{\pi}{4}x\right  )

    Therefore, F'\!\left(2\right)+F\!\left(2\right)=\sin\!\left(\  tfrac{\pi}{2}\right)+\left[\int_2^2\sin\!\left(\tfrac{\pi}{4}t\right)\,dt-5\right]=1-5=\color{red}\boxed{-4}

    Does this make sense?
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  4. #4
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    Yeah it makes sense now... Thanks you two alot.

    However, just to be sure, aren't integrals and derivatives inverses of each other? I was 99.9% sure they were.

    proof.
    f(x)=x^3
    f'(x)=3x^2
    \int 3x^2 dx=x^3 + C

    Am i right now? Or am i still Crazy?lol
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  5. #5
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    go see the discussion at this link ...

    http://http://www.physicsforums.com/...p/t-84539.html

    ... hope it will give you some insight. One must use the term "inverse" with extreme care.
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  6. #6
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    Quote Originally Posted by dandaman View Post
    Yeah it makes sense now... Thanks you two alot.

    However, just to be sure, aren't integrals and derivatives inverses of each other? I was 99.9% sure they were.

    proof.
    f(x)=x^3
    f'(x)=3x^2
    \int 3x^2 dx=x^3 + C

    Am i right now? Or am i still Crazy?lol
    Almost. Strictly speaking, two operations, f and g, are inverse if f(g(x))= x and g(f(x))= x. If you start with 3x^2, integrate it to get x^3+ C and then differentiate it, you get 3x^2 again. But if you do it the other way around, start with x^3, differentiate it to get 3x^2 and integrate, you get x^3+ C, NOT the same thing.

    The same sort of thing happens with functions: f(x)= x^2 and g(x)= \sqrt{x} are "almost" inverses. If you start with x= 2, square it to get 4 and then take the square root, you get 2 again. But if you start with x= -2, square it to get 4 and then take the squareroot, you get 2, not -2. Of course, to solve a quadratic equation, we take the square root anyway but remember that we need that " \pm".
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