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

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

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

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Answer 1 · 2018-02-21T17:05:44

The area is $\frac{1}{2}$ $1/2$ square units.

Explanation:

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

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

Let $e^{x} = t$ $e^x =t$ .

$t^{2} = 2 t$ $t^2 = 2t$

$t^{2} - 2 t = 0$ $t^2 - 2t = 0$

$t (t - 2) = 0$ $t(t - 2) = 0$

$t = 0 or 2$ $t = 0 or 2$

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

$x = \emptyset or ln 2$ $x = O/ or ln2$

Thus our interval will be $[0, ln 2]$ $[0, ln2]$ .

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

$A = \int_{0}^{ln 2} 2 e^{x} - e^{2 x} d x$ $A = int_0^(ln2) 2e^x -e^(2x)dx$

$A = [2 e^{x} - \frac{1}{2} e^{2 x})]_{0}^{ln 2}$ $A = [2e^x - 1/2e^(2x))]_0^(ln2)$

$A = 4 - 2 - (2 - \frac{1}{2} (1))$ $A = 4 - 2 - (2 - 1/2(1))$

$A = \frac{1}{2}$ $A = 1/2$ square units.

Hopefully this helps!

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

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

1 Answer

Explanation:

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Find the area of the region bounded by y=2e^xy=2exy=2e^x, y=e^(2x)y=e2xy=e^(2x) and x=0x=0x=0?

y=2e^xy=2exy=2e^x, y=e^(2x)y=e2xy=e^(2x) and x=0x=0x=0

1 Answer

Explanation:

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Impact of this question

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

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