Find the area of the region bounded by $y=2e^x$ , $y=e^(2x)$ and $x=0$ ?

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$y=2e^x$ , $y=e^(2x)$ and $x=0$

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Answer 1 · 2018-02-21T17:05:44

The area is $1/2$ square units.

You will want to find the intersection points of the curve in order to correctly sketch the region.

$e^(2x) = 2e^x$

Let $e^x =t$ .

$t^2 = 2t$

$t^2 - 2t = 0$

$t(t - 2) = 0$

$t = 0 or 2$

$e^x = 0 or e^x = 2$

$x = O/ or ln2$

Thus our interval will be $[0, ln2]$ .

We now note that on $[0, ln2]$ , the function $y = 2e^x$ has a larger value than $y = e^(2x)$ . Therefore, our integral will be

$A = int_0^(ln2) 2e^x -e^(2x)dx$

$A = [2e^x - 1/2e^(2x))]_0^(ln2)$

$A = 4 - 2 - (2 - 1/2(1))$

$A = 1/2$ square units.

Hopefully this helps!