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Can someone help me with this problem please? It says:

Find $k$ so that: $$0 < k < 5/2$$ and the area between $y=e^{2x}$, $y=e^{-2x}$ and $y=k$ results in $[ 10\ln(5/2) - 2/5 ]$.

I started looking for intersections between $y=k$ and those $2$ other curves and I got:
$x_1 = \frac{\ln(k)}{2}$
and
$x_2 = -\frac{\ln(k)}{2}$

After that i tried to find this:
$\int_{-\frac{\ln(k)}{2}}^0 k-e^{-2x} + \int_0^{\frac{\ln(k)}{2}}k-e^{2x}$

I'm stuck there...

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    So Integrate . . . – 2017-02-16
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    Yes i did that and i get 1 - k + k log(k)... but i don't know how to proceed with this 1 - k + k log(k) = 10ln(5/2) - 2/5 – 2017-02-16
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    My answer for the first is $$\int_{-\ln(k)/2}^{0}(k-e^{-2x})dx=kx+\frac12e^{-2x}\Big|_{-\ln(k)/2}^{0}=\frac12+\frac12k\ln k-\frac{k}{2}$$ Right.? – 2017-02-16
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    Yeah that's right, when yo do the other part the one that goes from 0 to ln(k)/2 you get the same... – 2017-02-16
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    And their addition.$1+k\ln k$.? – 2017-02-16
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    No, their adition to 1 + _k_ ln _k_ - _k_ – 2017-02-16
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    I solved $1+k\ln k-k=10\ln\dfrac52-\dfrac25$ with mathematica and answer was $k\sim7.57...$ so part of your problem is wrong. – 2017-02-16
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    the other issue there is that y=k curve can be above or under y=1 and at that point the 2 other curves intersect and the integration should change... – 2017-02-16
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    Can you write your problem in English, please? – 2017-02-16
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    In this case the answer is the same $1+k\ln k-k$. Not different $k$ where is. – 2017-02-16

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