How do you integrate (e^(x-2))^2/e^(2x)?

Oct 1, 2016

Use properties of exponents to rewrite.

Explanation:

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

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

Thus $\int {\left({e}^{x - 2}\right)}^{2} / {e}^{2 x} \mathrm{dx} = \int {e}^{-} 4 \mathrm{dx} = x {e}^{-} 4 + C$

Oct 1, 2016

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

Explanation:

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

$= \int {e}^{2 x} / \left({e}^{2 x} \cdot {e}^{4}\right) \mathrm{dx}$

$= \int \frac{\cancel{{e}^{2 x}}}{\cancel{{e}^{2 x}} {e}^{4}} \mathrm{dx}$

Thus the integral reduces to:

$\int \frac{1}{e} ^ 4 \mathrm{dx}$

Since $\frac{1}{e} ^ 4$ is a constant:

$= \frac{1}{e} ^ 4 \int \mathrm{dx} = \frac{1}{e} ^ 4 \cdot x + c$

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