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Math Help - Integral. Average Value.

  1. #1
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    Integral. Average Value.

    Integral. Average Value.-4d2q.pngIntegral. Average Value.-4d2a.png

    My question is about 4D-2. I see that the first step in the solutions manual is:

    \frac{1}{a} \int^{2a}_{a} \frac{1}{x} \;dx

    Why is there a \frac{1}{a} out front of the integral in that step?

    Thanks in advance for any help on this problem...
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  2. #2
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    Re: Integral. Average Value.

    Hey sepoto.

    The reason is for the 1/a is because that is the length of the interval. Remember that averages are defined to be average = total / length_of_interval. The total is the integral (since it is summing up everything) and the length is 2a - a = a.
    Thanks from sepoto
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