Integral of e^(5-2x) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 5 - 2 x$.

      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:

      $$- \frac{e^{5 - 2 x}}{2}$$

   Method #2

   1. Rewrite the integrand:

      $$e^{5 - 2 x} = e^{5} e^{- 2 x}$$

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

      $$\int e^{5} e^{- 2 x}\, dx = e^{5} \int e^{- 2 x}\, dx$$

      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: $- \frac{e^{5} e^{- 2 x}}{2}$

   Method #3

   1. Rewrite the integrand:

      $$e^{5 - 2 x} = e^{5} e^{- 2 x}$$

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

      $$\int e^{5} e^{- 2 x}\, dx = e^{5} \int e^{- 2 x}\, dx$$

      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: $- \frac{e^{5} e^{- 2 x}}{2}$

2. Add the constant of integration:

   $$- \frac{e^{5 - 2 x}}{2}+ \mathrm{constant}$$

The answer is:

$$- \frac{e^{5 - 2 x}}{2}+ \mathrm{constant}$$