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Math Help - Integration

  1. #1
    Member srirahulan's Avatar
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    Arrow Integration

    \int_{-2}^{2}|x+1|d(x)In this case can I break the mod value and write like this \int_{-2}^{2}xd(x)+\int_{-2}^{2}1d(x)
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  2. #2
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    Re: Integration

    No. On the interval [-2,1) |x+1| = -x-1, while on the interval [-1,2], |x+1| = x+1. So what you WANT is:

    -\int_{-2}^{-1}x\ dx - \int_{-2}^{-1}1\ dx + \int_{-1}^2 x\ dx + \int_{-1}^2 1\ dx
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  3. #3
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    Re: Integration

    Quote Originally Posted by Deveno View Post
    No. On the interval [-2,1) |x+1| = -x-1,
    TYPO! Deveno meant "On the interval [-2, -1].

    while on the interval [-1,2], |x+1| = x+1. So what you WANT is:

    -\int_{-2}^{-1}x\ dx - \int_{-2}^{-1}1\ dx + \int_{-1}^2 x\ dx + \int_{-1}^2 1\ dx
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  4. #4
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    Re: Integration

    Hello, srirahulan!

    \displaystyle\int_{\text{-}2}^{2}|x+1|\,dx

    \text{Graph the function: }\,y \:=\:|x+1|\,\text{ on }[\text{-}2,2]

    Code:
                      |
                      |          .*
                      |        .*.:
                      |      .*...:
                      |    .*.....:
                      |  .*.......:
                      |.*.........:
          *.         .*...........:
          :.*.     .*.|...........:
          :...*. .*...|...........:
       ---+-----*-----+-----+-----+---
         -2    -1     |     1     2
                      |
    \displaystyle\text{We want: }\:\int^{\text{-}1}_{\text{-}2}(\text{-}x-1)\,dx\:+\:\int^2_{\text{-}1}(x+1)\,dx
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  5. #5
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    Re: Integration

    Notice that, per Soroban's graph, this can be done with any "integration". The answer is the sum of the areas of two triangles.
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  6. #6
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    Re: Integration

    Indeed...using just "plain geometry" (ok, a lame pun), it is apparent that the integral:

    \int_{-2}^2 x+1\ dx = 4, while:

    \int_{-2}^2 |x+1|\ dx = 5.

    The reason being, the triangle with base extending from (-2,0) to (-1,0) in the former integral has a "peak" at (-2,-1) and lies BELOW the x-axis (so has "negative area", that is the integral is negative in this region), while in the latter integral lies above the x-axis (with "peak" at (-2,1)). This triangular area has a base of 1, and a height of 1, and so has area 1/2, making a "swing" of a difference of 1 between the two integrals.

    And yes, Halls, that WAS a typo. :P
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