Integral of e^(-2x) dx

1. Let $u = - 2 x$.

   Then let $du = - 2 dx$ and substitute $- \frac{du}{2}$:

   $$\int \left(- \frac{e^{u}}{2}\right)\, du$$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\text{False}$$

      1. The integral of the exponential function is itself.

         $$\int e^{u}\, du = e^{u}$$

      So, the result is: $- \frac{e^{u}}{2}$

   Now substitute $u$ back in:

   $$- \frac{e^{- 2 x}}{2}$$

2. Add the constant of integration:

   $$- \frac{e^{- 2 x}}{2}+ \mathrm{constant}$$

The answer is:

$$- \frac{e^{- 2 x}}{2}+ \mathrm{constant}$$