The area bounded by the X-axis and the curve | vedclass

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(A) For the $X$-axis,$y=0$.Therefore,$x(x-2)(x+1)=0$,which gives $x=0, x=2, x=-1$.The required area is given by the integral of $|y|$ with respect to

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$\frac{37}{12}$ sq. units

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(A) For the $X$-axis,$y=0$.Therefore,$x(x-2)(x+1)=0$,which gives $x=0, x=2, x=-1$.The required area is given by the integral of $|y|$ with respect to $x$ from $-1$ to $2$.$	ext{Area} = \int_{-1}^0 y \, dx + \left| \int_0^2 y \, dx ight|$$	ext{Area} = \int_{-1}^0 (x^3-x^2-2x) \, dx + \left| \int_0^2 (x^3-x^2-2x) \, dx ight|$Evaluating the integrals:$\int (x^3-x^2-2x) \, dx = \frac{x^4}{4} - \frac{x^3}{3} - x^2$For the interval $[-1, 0]$:$\left[ \frac{x^4}{4} - \frac{x^3}{3} - x^2 ight]_{-1}^0 = 0 - \left( \frac{1}{4} - \frac{-1}{3} - 1 ight) = - \left( \frac{3+4-12}{12} ight) = - \left( \frac{-5}{12} ight) = \frac{5}{12}$For the interval $[0, 2]$:$\left[ \frac{x^4}{4} - \frac{x^3}{3} - x^2 ight]_0^2 = \left( \frac{16}{4} - \frac{8}{3} - 4 ight) - 0 = 4 - \frac{8}{3} - 4 = -\frac{8}{3}$Taking the absolute value,we get $|-\frac{8}{3}| = \frac{8}{3}$.Total Area = $\frac{5}{12} + \frac{8}{3} = \frac{5+32}{12} = \frac{37}{12}$ sq. units.

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

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