Integral of e^(-10x) dx

1. Let $u = - 10 x$.

   Then let $du = - 10 dx$ and substitute $- \frac{du}{10}$:

   $$\int \left(- \frac{e^{u}}{10}\right)\, du$$

   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: $- \frac{e^{u}}{10}$

   Now substitute $u$ back in:

   $$- \frac{e^{- 10 x}}{10}$$

2. Add the constant of integration:

   $$- \frac{e^{- 10 x}}{10}+ \mathrm{constant}$$

The answer is:

$$- \frac{e^{- 10 x}}{10}+ \mathrm{constant}$$