Integral of $$$\left(20 - 5 x\right)^{2}$$$

Find $$$\int \left(20 - 5 x\right)^{2}\, dx$$$.

Let $$$u=20 - 5 x$$$.

Then $$$du=\left(20 - 5 x\right)^{\prime }dx = - 5 dx$$$ (steps can be seen »), and we have that $$$dx = - \frac{du}{5}$$$.

So,

$${\color{red}{\int{\left(20 - 5 x\right)^{2} d x}}} = {\color{red}{\int{\left(- \frac{u^{2}}{5}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=- \frac{1}{5}$$$ and $$$f{\left(u \right)} = u^{2}$$$:

$${\color{red}{\int{\left(- \frac{u^{2}}{5}\right)d u}}} = {\color{red}{\left(- \frac{\int{u^{2} d u}}{5}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:

$$- \frac{{\color{red}{\int{u^{2} d u}}}}{5}=- \frac{{\color{red}{\frac{u^{1 + 2}}{1 + 2}}}}{5}=- \frac{{\color{red}{\left(\frac{u^{3}}{3}\right)}}}{5}$$

Recall that $$$u=20 - 5 x$$$:

$$- \frac{{\color{red}{u}}^{3}}{15} = - \frac{{\color{red}{\left(20 - 5 x\right)}}^{3}}{15}$$

Therefore,

$$\int{\left(20 - 5 x\right)^{2} d x} = - \frac{\left(20 - 5 x\right)^{3}}{15}$$

Simplify:

$$\int{\left(20 - 5 x\right)^{2} d x} = \frac{25 \left(x - 4\right)^{3}}{3}$$

Add the constant of integration:

$$\int{\left(20 - 5 x\right)^{2} d x} = \frac{25 \left(x - 4\right)^{3}}{3}+C$$

$$$\int \left(20 - 5 x\right)^{2}\, dx = \frac{25 \left(x - 4\right)^{3}}{3} + C$$$A