How do you integrate $\frac{{(e^{x - 2})}^{2}}{e^{2 x}}$ $(e^(x-2))^2/e^(2x)$ ?

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Answer 1 · 2016-10-01T02:50:41

Use properties of exponents to rewrite.

${(e^{x - 2})}^{2} = e^{2 (x - 2)} = e^{2 x - 4} = e^{2 x} e^{- 4}$ $(e^(x-2))^2 = e^(2(x-2)) = e^(2x-4) = e^(2x)e^(-4)$

So $\frac{{(e^{x - 2})}^{2}}{e^{2 x}} = \frac{e^{2 x} e^{- 4}}{e^{2 x}} = e^{- 4}$ $(e^(x-2))^2/e^(2x) = (e^(2x)e^-4)/e^(2x) = e^-4$

Thus $\int \frac{{(e^{x - 2})}^{2}}{e^{2 x}} d x = \int e^{- 4} d x = x e^{- 4} + C$ $int (e^(x-2))^2/e^(2x) dx = int e^-4 dx = x e^-4 +C$

Answer 2 · 2016-10-01T02:59:48

$\frac{x}{e^{4}} + c$ $x/e^4 + c$

$\int \frac{{(e^{x - 2})}^{2}}{e^{2 x}} d x = \int \frac{e^{2 x - 4}}{e^{2 x}} d x$ $int(e^(x-2))^2/e^(2x) dx = int(e^(2x-4))/e^(2x) dx$

$= \int \frac{e^{2 x}}{e^{2 x} \cdot e^{4}} d x$ $= inte^(2x)/(e^(2x)*e^4) dx$

$= int cancel(e^(2x))/(cancel(e^(2x))e^4) dx$

Thus the integral reduces to:

$int 1/e^4 dx$

Since $1/e^4$ is a constant:

$= 1/e^4 int dx = 1/e^4*x +c$

$=x/e^4 + c$

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