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Thread: Definite Integral

  1. #1
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    Definite Integral

    If the definite integral from -2 to 6 of f(x)dx=10 and the definite integral from 2 to 6 of f(x)dx=3, then the definite integral from 2 to 6 of f(x-4)dx= ?

    I don't understand how to solve definite integrals when the function has something more than just x inside the parenthesis such as f(4-x).
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    Quote Originally Posted by frog09 View Post
    If the definite integral from -2 to 6 of f(x)dx=10 and the definite integral from 2 to 6 of f(x)dx=3, then the definite integral from 2 to 6 of f(x-4)dx= ?

    I don't understand how to solve definite integrals when the function has something more than just x inside the parenthesis such as f(4-x).
    Here is a hint: what does substituting $\displaystyle 4-x$ instead of $\displaystyle x$ do to the function? An easy example to find out is to graph $\displaystyle f(x) = x^2$ and then graph $\displaystyle f(x-4) = (x-4)^2$. Compare $\displaystyle \int_2^6 x^2 dx $ and $\displaystyle \int_6^{10} (x-4)^2 dx$

    Do you see why? What I used a change of variables for the second integral, where $\displaystyle y=x-4,y+4=x$ and $\displaystyle dy = dx$?

    Once you figure that out, recall that for that integral is a linear operator:
    $\displaystyle \int_a^c f(x) dx = \int_a^b f(x) dx + \int_b^c f(x) dx$ where $\displaystyle a<b<c \in \mathbb{R}$

    You should get the answer to be 7: $\displaystyle \int_2^6 f(x-4) dx = 7$
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  3. #3
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    Quote Originally Posted by frog09 View Post
    If the definite integral from -2 to 6 of f(x)dx=10 and the definite integral from 2 to 6 of f(x)dx=3, then the definite integral from 2 to 6 of f(x-4)dx= ?

    I don't understand how to solve definite integrals when the function has something more than just x inside the parenthesis such as f(4-x).
    Hint: Let $\displaystyle t = 4-x$ in $\displaystyle \int^6_2 f(4-x)\;dx.$ (Too slow)
    Last edited by Jester; Jan 2nd 2009 at 07:52 AM. Reason: too slow
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  4. #4
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    Quote Originally Posted by frog09 View Post
    If the definite integral from -2 to 6 of f(x)dx=10 and the definite integral from 2 to 6 of f(x)dx=3, then the definite integral from 2 to 6 of f(x-4)dx= ?

    I don't understand how to solve definite integrals when the function has something more than just x inside the parenthesis such as f(4-x).
    $\displaystyle \int_{-2}^6 f(x) \, dx = 10$

    $\displaystyle \int_2^6 f(x) \, dx = 3$

    $\displaystyle \int_2^6 f(x-4) \, dx = \, ?$

    use the method of substitution ...

    $\displaystyle u = x - 4$

    $\displaystyle du = dx$

    substitute and reset the limits of integration ...

    $\displaystyle \int_2^6 f(x-4) \, dx$

    $\displaystyle \int_{-2}^2 f(u) \, du$


    $\displaystyle \int_{-2}^6 f(x) \, dx - \int_2^6 f(x) \, dx = \int_{-2}^2 f(x) \, dx$

    $\displaystyle 10 - 3 = 7$

    since $\displaystyle \int_{-2}^2 f(u) \, du = \int_{-2}^2 f(x) \, dx$

    $\displaystyle \int_{-2}^2 f(u) \, du = \int_2^6 f(x-4) \, dx = 7$
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    LOL @ danny arrigo

    Sorry
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    I was re-checking this problem and realized I copied down the last part wrong. It is really supposed to be:
    $\displaystyle
    \int_2^6 f(4-x) , dx = ?
    $

    i substitute u=4-x and get -du=dx. and get the new limits as
    $\displaystyle
    \int_2^{-2} f(u) du
    $

    which would be
    $\displaystyle
    -\int_{-2}^2 f(u) du
    $

    does that -du change the problem? or would it still be 7?
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  7. #7
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    Quote Originally Posted by frog09 View Post
    I was re-checking this problem and realized I copied down the last part wrong. It is really supposed to be:
    $\displaystyle
    \int_2^6 f(4-x) , dx = ?
    $

    i substitute u=4-x and get -du=dx. and get the new limits as
    $\displaystyle
    \int_2^{-2} f(u) du
    $

    which would be
    $\displaystyle
    -\int_{-2}^2 f(u) du
    $

    does that -du change the problem? or would it still be 7?
    You would get a negative from the dx so your second last line should be - and the last line should be +.
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