<s>[INST] 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:
$\int_{-2}^2 (1-2x)^2dx$
$=\int_{-2}^2 1-4x+4x^2dx$
$=(x-\frac{4}{2}x^2+\frac{4}{3}x^3) |_{-2}^2$
$=(x-2x^2+\frac{4}{3}x^3) |_{-2}^2$
$=(2-2(2)^2+\frac{4}{3}(2)^3)-(-2-2(-2)^2+\frac{4}{3}(-2)^3)$
$=(2-8+\frac{32}{3})-(-2-8+\frac{-32}{3})$
$=2-8+\frac{32}{3}+2+8+\frac{32}{3})$
$=4+\frac{64}{3}$
$\frac{12}{3}+\frac{64}{3}$
$=\frac{76}{3}$
 [/INST] $$
\int_{-2}^{2}{(1-2x)^{2}dx}\\=\int_{-2}^{2}{1-4x+4x^2dx}\\=\int_{-2}^{2}{1dx}+\int_{-2}^{2}{4xdx}+\int_{-2}^{2}{4x^2dx}
$$
Now lets solve those integrals.
$$
\int_{-2}^{2}{1dx}=4\\
$$
since it has only a constant
$$
\int_{-2}^{2}{4xdx}=0\\
$$
since it's an even function.
$$
\int_{-2}^{2}{4x^2}\\
=4\times\int_{-2}^{2}{x^2}\\
=4[\frac{x^3}{3}]_{-2}^{2}\\
=4\times(\frac{8}{3}-\frac{-8}{3})\\
=4\times\frac{16}{3}\\
=\frac{64}{3}
$$
so
$$
=4-0+\frac{64}{3}
=\frac{76}{3}
$$
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