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• Notes

Find the area of the region bounded by the graphs of the equation $: y=\sqrt{x}, y=x^{4}$

$y=x \rightarrow \sqrt{x}=x^{4}$

$\longrightarrow(\sqrt{x})^{2}=\left(x^{4}\right)^{2}$

$x=x^{8} \rightarrow x^{8}-x=0 \rightarrow x\left(x^{7}-1\right)=0$

$x=0 \quad \quad x=1$

$A=\int_{0}^{1}\left(\sqrt{x}-x^{4}\right) d x$

$A=\left[\frac{x^{3 / 2}}{3 / 2}-\frac{x^{5}}{5}\right]=\frac{2}{3}-\frac{1}{5}=\frac{7}{15}$

Find the area of the region bounded by the graphs of the equation $: y+x^{2}-2 x=0$ and $y+2 x=0$

$y+x^{2}-2 x=0 \rightarrow y=2 x-x^{2}$

$y+2 x=0 \longrightarrow y=-2 x$

$y=y \rightarrow 2 x-x^{2}=-2 x \rightarrow 4 x-x^{2}=0 \rightarrow x(4-x)=0$

$x=0 \quad \quad x=4$

Area $=\int_{0}^{4} 2 x-x^{2}-(-2 x) d x=\int_{0}^{4} 2 x-x^{2}+2 x d x$

$=\int_{0}^{4}-x^{2}+4 x d x=\left[\frac{-x^{3}}{3}+2 x^{2}\right]_{0}^{4}=32-\frac{64}{3}$

$=\frac{32}{3}$

Find the area of the region bounded by $x$ -axis and $y=x^{3}-9 x$

$y=0$

$y=x^{3}-9 x$

$x(x-3)(x+3)=0$

$x=0 \quad x=3 \quad x=-3$

$\text { For } A_{1} : A_{1}=\int_{-3}^{0}\left(x^{3}-9 x-0\right) d x=\left[\frac{x^{4}}{4}-\frac{9 x^{2}}{2}\right]_{-3}^{0}=0-\left(\frac{(-3)^{4}}{4}-\frac{9(-3)^{2}}{2}\right)$

$=\frac{81}{4}$

$\text { For } A_{2}=\int_{0}^{3} 0-\left(x^{3}-9 x\right) d x=-\left[\frac{x^{4}}{4}-\frac{9 x^{2}}{2}\right]_{0}^{3}=\frac{81}{4}$

$A_T=A_{1}+A_{2}=\frac{81}{4}+\frac{81}{4}=\frac{81}{2}$