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Math Help - i dunno how to do these integral problems

  1. #1
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    i dunno how to do these integral problems

    there are 2 that i dunno how to do.
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  2. #2
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    Re:

    Re:
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  3. #3
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    for the 1st problem

    for the 1st problem
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  4. #4
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    Hello, falloutdude!

    Did you make a sketch for the second problem?


    Let: . f(x) \;=\;\begin{Bmatrix} 0 & \text{if }x< 0 \\<br />
x & \text{if }0 \leq x\leq 3 \\<br />
6-x & \text{if }3< x\leq 6 \\<br />
0 & \text{if }x > 6\end{Bmatrix} . and: . g(x) \:=\:\int^x_0f(t)\,dx

    Find an expression for g(x) when  3 < x  <6.
    The graph looks like this:
    Code:
                  |
                  |       (3,3)
                  |         *
                  |       *:|:*
                  |     *:::|:::*
                  |   *:::::|:::| *
                  | *:::::::|:::|   *
        - - * * * * - - - - + - + - - * * * * - -
                  0         3   x     6
    g(x) is the area under the graph from 0 to x.

    The left region is a right triangle with base 3 and height 3; its area is \frac{9}{2}

    The right region is the area under y \:=\:6 - x from 3 to x.
    . . This is: . \int^x_3(6 - x)\,dx \;=\;6x-\frac{1}{2}x^2\bigg]^x_3 \;=\;6x - \frac{1}{2}x^2 - \frac{27}{2}


    Therefore: . g(x) \;=\;\frac{9}{2} + \left(6x -\frac{1}{2}x^3 - \frac{27}{2}\right)\quad\Rightarrow\quad\boxed{g(x  )\;=\;6x - \frac{1}{2}x^2 - 9}

    Last edited by Soroban; June 15th 2007 at 08:52 PM.
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  5. #5
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    the graph of f(x)

    Quote Originally Posted by Soroban View Post
    Hello, falloutdude!

    Did you make a sketch for the second problem?

    The graph looks like this:
    Code:
                  |
                  |       (3,3)
                  |         *
                  |       *:|:*
                  |     *:::|:::*
                  |   *:::::|:::| *
                  | *:::::::|:::|   *
        - - * * * * - - - - + - + - - * * * * - -
                  0         3   x     6
    g(x) is the area under the graph from 0 to x.

    The left region is a right triangle with base 3 and height 3; its area is \frac{9}{2}

    The right region is the area under y \:=\:6 - x from 3 to x.
    . . This is: . \int^x_3(6 - x)\,dx \;=\;6x-\frac{1}{2}x^2\bigg]^x_3 \;=\;6x - \frac{1}{2}x^2 - \frac{27}{2}


    Therefore: . g(x) \;=\;\frac{9}{2} + \left(6x -\frac{1}{2}x^3 - \frac{27}{2}\right)\quad\Rightarrow\quad\boxed{g(x  )\;=\;6x^2 - \frac{1}{2}x^2 - 9}
    the graph of f(x)
    Attached Thumbnails Attached Thumbnails i dunno how to do these integral problems-23june2007-f-x-.gif  
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  6. #6
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    and the graph of g(x)

    and the graph of g(x)
    Attached Thumbnails Attached Thumbnails i dunno how to do these integral problems-23june2007-g-x-.gif  
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