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Math Help - Function integrate

  1. #1
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    Question Function integrate

    I have to find the value of:

    INTEGRATION[ f'(x) ((f(x))^3 -3f(x)-5) dx ]

    with upper boud 6 and lower bound 4

    given that
    f(4) = 1, f '(4) = -7, f(6) = 2 and f '(6) = 2

    So i tried the following which i dont think is right?
    2((2)^3 - 3 * 2 - 5) - (-7) - 3 * 1 - 5)
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by taurus View Post
    I have to find the value of:

    INTEGRATION[ f'(x) ((f(x))^3 -3f(x)-5) dx ]

    with upper boud 6 and lower bound 4

    given that
    f(4) = 1, f '(4) = -7, f(6) = 2 and f '(6) = 2

    So i tried the following which i dont think is right?
    2((2)^3 - 3 * 2 - 5) - (-7) - 3 * 1 - 5)
    Observe:

    \frac{d}{dx}\left[\frac{(f(x))^4}{4} -\frac{3f(x)^2}{2}-5f(x)+C\right]=f'(x) [(f(x))^3 -3f(x)-5]

    RonL
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  3. #3
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    i dont get it?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by taurus View Post
    i dont get it?
    Your integrand is easily seen to be the derivative of the
    given polynomial in f(x). Then the Fundamental
    Theorem of Calculus tells us that the integral is that
    polynomial in f(x).

    That is you use the result that:

    \int \frac{d}{dx}g(x) ~dx=g(x)+C

    or for a definite integral:

    \int_a^b \frac{d}{dx}g(x) ~dx=g(b)-g(a)

    RonL
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  5. #5
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    Would the answer be: -5456.83
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  6. #6
    Super Member wingless's Avatar
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    \int^{6}_{4}f'(x)[f^3(x)-3f(x)-5]~dx

    \int^{6}_{4}f^3(x)f'(x)-3f(x)f'(x)-5f'(x)~dx

    \int^{6}_{4}f^3(x)f'(x)~dx-3\int^{6}_{4}f(x)f'(x)~dx-5\int^{6}_{4}f'(x)~dx

    Use f'(x)~dx = d(f(x))

    \int^{6}_{4}f^3(x)~d(f(x))-3\int^{6}_{4}f(x)~d(f(x))-5\int^{6}_{4}d(f(x))

    \frac{f^4(x)}{4} - \frac{3f^2(x)}{2} - 5f(x) ~\displaystyle{{|}^{6}_{4}}

    Now go on..
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