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Math Help - 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 4-x instead of x do to the function? An easy example to find out is to graph f(x) = x^2 and then graph f(x-4) = (x-4)^2. Compare \int_2^6 x^2 dx and \int_6^{10} (x-4)^2 dx

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

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

    You should get the answer to be 7: \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  t = 4-x in \int^6_2 f(4-x)\;dx. (Too slow)
    Last edited by Jester; January 2nd 2009 at 07:52 AM. Reason: too slow
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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).
    \int_{-2}^6 f(x) \, dx = 10

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

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

    use the method of substitution ...

    u = x - 4

    du = dx

    substitute and reset the limits of integration ...

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

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


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

    10 - 3 = 7

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

    \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:
    <br />
\int_2^6 f(4-x) , dx = ?<br />

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

    which would be
    <br />
-\int_{-2}^2 f(u) du<br />

    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:
    <br />
\int_2^6 f(4-x) , dx = ?<br />

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

    which would be
    <br />
-\int_{-2}^2 f(u) du<br />

    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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