Evaluate integral of e to the power of(minus two x) with respect to x

$\int e^{- 2 x} d x$

Let $u = - 2 x$ . Then $d u = - 2 d x$ , so $- \frac{1}{2} d u = d x$ . Should be rewritten using $u$ and $d$ $u$ .

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Let $u = - 2 x$ . Find $\frac{d u}{d x}$ .

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Rewrite.

$\frac{- 2}{1}$

First, divide $- 2$ by $1$ .

$- 2$

Rewrite the problem using $u$ and $d u$ .

$\int e^{u} \frac{1}{- 2} d u$

Simplify.

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Put the negative in front of the fraction.

$\int e^{u} (- \frac{1}{2}) d u$

Add $e^{u}$ and $\frac{1}{2}$ .

$\int - \frac{e^{u}}{2} d u$

Since $- 1$ is constant with respect to $u$ , move $- 1$ out of the integral.

$- \int \frac{e^{u}}{2} d u$

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$- (\frac{1}{2} \int e^{u} d u)$

The integral of $e^{u}$ with respect to $u$ is $e^{u}$ .

$- \frac{1}{2} (e^{u} + C)$

Simplify.

$- \frac{1}{2} e^{u} + C$

Replace all occurrences of $u$ with $- 2 x$ .

$- \frac{1}{2} e^{- 2 x} + C$

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