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

$$y=0$$

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

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

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

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

$$=\frac{81}{4}$$

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

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

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