Integral of e^-t dx

1. Let $u = - t$.

   Then let $du = - dt$ and substitute $- du$:

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

   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: $- e^{u}$

   Now substitute $u$ back in:

   $$- e^{- t}$$

2. Add the constant of integration:

   $$- e^{- t}+ \mathrm{constant}$$

The answer is:

$$- e^{- t}+ \mathrm{constant}$$