Integral of sqrt(8-2x)dx dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 8 - 2 x$.

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

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

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

         $$\int \sqrt{u}\, du = - \frac{\int \sqrt{u}\, du}{2}$$

         1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

            $$\int \sqrt{u}\, du = \frac{2 u^{\frac{3}{2}}}{3}$$

         So, the result is: $- \frac{u^{\frac{3}{2}}}{3}$

      Now substitute $u$ back in:

      $$- \frac{\left(8 - 2 x\right)^{\frac{3}{2}}}{3}$$

   Method #2

   1. Rewrite the integrand:

      $$\sqrt{8 - 2 x} = \sqrt{2} \sqrt{4 - x}$$

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

      $$\int \sqrt{2} \sqrt{4 - x}\, dx = \sqrt{2} \int \sqrt{4 - x}\, dx$$

      1. Let $u = 4 - x$.

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

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

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

            $$\int \sqrt{u}\, du = - \int \sqrt{u}\, du$$

            1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

               $$\int \sqrt{u}\, du = \frac{2 u^{\frac{3}{2}}}{3}$$

            So, the result is: $- \frac{2 u^{\frac{3}{2}}}{3}$

         Now substitute $u$ back in:

         $$- \frac{2 \left(4 - x\right)^{\frac{3}{2}}}{3}$$

      So, the result is: $- \frac{2 \sqrt{2} \left(4 - x\right)^{\frac{3}{2}}}{3}$

2. Now simplify:

   $$- \frac{2 \sqrt{2} \left(4 - x\right)^{\frac{3}{2}}}{3}$$

3. Add the constant of integration:

   $$- \frac{2 \sqrt{2} \left(4 - x\right)^{\frac{3}{2}}}{3}+ \mathrm{constant}$$

The answer is:

$$- \frac{2 \sqrt{2} \left(4 - x\right)^{\frac{3}{2}}}{3}+ \mathrm{constant}$$