The area bounded by the X-axis and the curve $y=x(x-2)(x+1)$ is

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Aicharya · Accepted Answer

$\begin{array}{ll} & 	ext { For X-axis, } \ & y=0 \ 	herefore \quad & x(x-2)(x+1)=0 \ & \Rightarrow x=0 	ext { or } x=2 	ext { or } x=-1\end{array}$ $\begin{aligned} & 	ext { Required area }=\int_{-1}^0 y \mathrm{~d} x+\left|\int_0^2 y \mathrm{~d} xight| \ & =\int_{-1}^0 x(x-2)(x+1) \mathrm{d} x+\left|\int_0^2 x(x-2)(x+1) \mathrm{d} xight| \ & =\int_{-1}^0\left(x^3-x^2-2 xight) \mathrm{d} x+\left|\int_0^2\left(x^3-x^2-2 xight) \mathrm{d} xight| \ & =\left[\frac{x^4}{4}-\frac{x^3}{3}-x^2ight]_{-1}^0+\left|\left[\frac{x^4}{4}-\frac{x^3}{3}-x^2ight]_0^2ight| \ & =\frac{5}{12}+\left|-\frac{8}{3}ight| \ & =\frac{37}{12} 	ext { sq. units }\end{aligned}$

The area bounded by the X-axis and the curve $y=x(x-2)(x+1)$ is

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