Find the area under the graph of the function $f(x)=(1-2x)^2$ between $x=-2$ and $x=2$

Question

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 $x=2$

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The area under the graph of $f(x)$ between $x=-2$ and $x=2$ is exactly equal to:

$\int_{-2}^2 (1-2x)^2dx$

$=\int_{-2}^2 1-4x+4x^2dx$

$=(x-\frac{4}{2}x^2+\frac{4}{3}x^3) |_{-2}^2$

$=(x-2x^2+\frac{4}{3}x^3) |_{-2}^2$

$=(2-2(2)^2+\frac{4}{3}(2)^3)-(-2-2(-2)^2+\frac{4}{3}(-2)^3)$

$=(2-8+\frac{32}{3})-(-2-8+\frac{-32}{3})$

$=2-8+\frac{32}{3}+2+8+\frac{32}{3})$

$=4+\frac{64}{3}$

$\frac{12}{3}+\frac{64}{3}$

$=\frac{76}{3}$

Answer 1

$$ \int_{-2}^{2}{(1-2x)^{2}dx}\\=\int_{-2}^{2}{1-4x+4x^2dx}\\=\int_{-2}^{2}{1dx}+\int_{-2}^{2}{4xdx}+\int_{-2}^{2}{4x^2dx} $$ 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 $$ =4-0+\frac{64}{3} =\frac{76}{3} $$

Answer 2

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