Find the area under the graph of the function $f(x)=(1-2x)^2$ between $x=-2$ and $x=2$ 5.4
Can somebody verify this solution for me?
Find the area under the graph of the function $f(x)=(1-2x)^2$ between $x=-2$ and $x=2$

The area under the graph of $f(x)$ between $x=-2$ and $x=2$ is exactly equal to:
$\int_{-2}^2 (1-2x)^2dx$
$=\int_{-2}^2 1-4x+4x^2dx$
$=(x-\frac{4}{2}x^2+\frac{4}{3}x^3) |_{-2}^2$
$=(x-2x^2+\frac{4}{3}x^3) |_{-2}^2$
$=(2-2(2)^2+\frac{4}{3}(2)^3)-(-2-2(-2)^2+\frac{4}{3}(-2)^3)$
$=(2-8+\frac{32}{3})-(-2-8+\frac{-32}{3})$
$=2-8+\frac{32}{3}+2+8+\frac{32}{3})$
$=4+\frac{64}{3}$
$\frac{12}{3}+\frac{64}{3}$
$=\frac{76}{3}$

You would spare your time if you make a substitution $ t=2x-1$. $$\int_{-2}^2 (1-2x)^2dx  = {1\over 2}\int_{-5}^3 t^2dt =...$$
