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Math Help - Sketching the Graph of an Interval (Help Please)

  1. #1
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    Sketching the Graph of an Interval (Help Please)

    Given first sketch the graph y = f(x) on the given interval.
    Then find the integral of
    f using your knowledge of area formulas for

    rectangles, triangles and circles.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by evansf View Post
    Given first sketch the graph y = f(x) on the given interval.

    Then find the integral of
    f using your knowledge of area formulas for
    rectangles, triangles and circles.

    \int_{-2}^4 |3 - |3x||dx = \int_{-2}^4 |3 - 3|x||dx = \int_{-2}^4 3|1 - |x||dx.. now, this can be separated to

    \int_{-2}^4 |3 - |3x||dx = ... = \int_{-2}^0 3|1 - |x||dx + \int_{0}^4 3|1 - |x||dx..

    we know, in the first addend, x is negative, hence we take |x| = -x; on the other hand, in the second addend, x is positive hence |x| = x

    so the integral would be

    \int_{-2}^0 3|1 - |x||dx + \int_{0}^4 3|1 - |x||dx = \int_{-2}^0 3|1 + x|dx + \int_{0}^4 3|1 - x|dx

    now, we have to check where |1-x| is positive or negative and |1+x| is positive or negative..


    for |1-x|, we take 1-x if 1-x >= 0 or x <= 1; we take x-1 otherwise

    for |1+x|, we take 1+x if 1+x >= 0 or x >= -1; we take -x-1 otherwise

    so now,

    ... = \int_{-2}^{-1} 3(-1 - x)dx + \int_{-1}^{0} 3(1 + x)dx + \int_{0}^1 3(1 - x)dx + \int_{1}^4 3(x - 1)dx

    so can you do it now??
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    How do you integrate it?

    ...
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  4. #4
    MHF Contributor kalagota's Avatar
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    they are just linear functions, then they can easily be integrated..
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by evansf View Post
    ...
    You might try posting a message in the message box, rather than just in the title.

    Take a look at your graph. The integral is the total area between the function and the x axis. (Areas under the x axis count as negative.) As your function is composed of sections of straight lines, your areas are triangles. Easy to find the area of these.

    -Dan
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