Integral of 2e^(-2x) dx

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

   $$\int 2 e^{- 2 x}\, dx = 2 \int e^{- 2 x}\, dx$$

   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:

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

   So, the result is: $- e^{- 2 x}$

2. Add the constant of integration:

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

The answer is:

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