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Thread: 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:

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

    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 $\displaystyle f(x)$. Then the Fundamental
    Theorem of Calculus tells us that the integral is that
    polynomial in $\displaystyle f(x)$.

    That is you use the result that:

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

    or for a definite integral:

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

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

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

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

    Use $\displaystyle f'(x)~dx = d(f(x))$

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

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

    Now go on..
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