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Math Help - find the area bounded by the y-axis

  1. #1
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    find the area bounded by the y-axis

    from the graphs

    y = e^x

    y = 7-e^x

    i got the lines in a online function grapher but not sure where the area is bounded by the y-axis and which intervals i should use.
    here is the graph with both functions.

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  2. #2
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    I think the region you're supposed to evaluate starts at y=1, has a right edge where the green and red graphs intersect, and an upper bound where the green graph crosses the y-axis again, which should be at y=6.

    So because of the way the region is set up, we should integrate with respect to y instead of x. That means we need to solve both equations for x.

    (y=e^x) \rightarrow (x=\ln(y))
    (y=7-e^x) \rightarrow (x=\ln(7-y))

    So looking at the region it looks like it's good to break it up into two smaller regions. From y=1 to where the graphs intersect, the x-values of the region are defined by x=\ln(y) and after they intersect until y=6 the x-values are defined by x=\ln(7-y)

    So let's figure out where they intersect.

    \ln(y)=\ln(7-y)
    y=7-y
    y=\frac{7}{2}

    So I think the whole are is:

    \int_{1}^{\frac{7}{2}} \ln(y)dy + \int_{\frac{7}{2}}^{6} \ln(7-y)dy

    Does that make sense? Hopefully I don't have any huge errors.
    Last edited by Jameson; November 3rd 2008 at 08:11 PM. Reason: mistake
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  3. #3
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    Quote Originally Posted by Jameson View Post
    I think the region you're supposed to evaluate starts at y=1, has a right edge where the green and red graphs intersect, and an upper bound where the green graph crosses the y-axis again, which should be at y=7.

    So because of the way the region is set up, we should integrate with respect to y instead of x. That means we need to solve both equations for x.

    (y=e^x) \rightarrow (x=\ln(y))
    (y=7-e^x) \rightarrow (x=\ln(7-y))

    So looking at the region it looks like it's good to break it up into two smaller regions. From y=1 to where the graphs intersect, the x-values of the region are defined by x=\ln(y) and after they intersect until y=7 the x-values are defined by x=\ln(7-y)

    So let's figure out where they intersect.

    \ln(y)=\ln(7-y)
    y=7-y
    y=\frac{7}{2}

    So I think the whole are is:

    \int_{1}^{\frac{7}{2}} \ln(y)dy + \int_{\frac{7}{2}}^{7} \ln(7-y)dy

    Does that make sense? Hopefully I don't have any huge errors.
    just letting you know i shrunk the graph and the green line crosses the y axis on 6, would that make any major changes to the working out? just change the interval number yeah? so from 7/2 till 6? makes more sence now man. thanks alot!

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  4. #4
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    Doh! e^0 = 1, not 0. Brain fade. That shouldn't change anything though.
    Last edited by mr fantastic; December 8th 2009 at 11:58 AM.
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