# Determine the area of the region bounded by $y=2e^x$, $y=e^{2x}$ and $x=0$

$$y_1 = 2e^x$$ $$y_2 = e^{2x}$$ $$x=0$$

I was thinking of finding the $x$-intercepts first, so $2e^x= e^{2x}$.

What is next?

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Think of the equation as:$$2e^x=(e^x)^2.$$Let $z=e^x$. What is the resulting equation? How would you solve it?

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thank you! it helped me a lot actually. –  Anaïs De Potesta Apr 29 '13 at 2:06

I assume you already know Double Integration. If you don't, wait for someone else's answer. Once you find the intercept, call it $z$,

Area = $\displaystyle \int_{x=0}^{x=z}\int_{y=e^{2x}}^{y=2e^x}dy.dx$

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