The area (in sq. text{ units}) bounded by the | vedclass

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(A) Given the curve $y = x|x|$.Since $x|x|$ is an odd function,we analyze the integral $\int_{-1}^{1} x|x| dx$.However,the question asks for the area

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$2/3$

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(A) Given the curve $y = x|x|$.Since $x|x|$ is an odd function,we analyze the integral $\int_{-1}^{1} x|x| dx$.However,the question asks for the area bounded by the curve and the $X$-axis,which is given by $\int_{-1}^{1} |y| dx = \int_{-1}^{1} |x|x|| dx = \int_{-1}^{1} |x|^2 dx = \int_{-1}^{1} x^2 dx$.$	ext{Area} = \int_{-1}^{1} x^2 dx = 2 \int_{0}^{1} x^2 dx$.$= 2 \left[ \frac{x^3}{3} ight]_{0}^{1}$.$= 2 	imes \left( \frac{1}{3} - 0 ight) = \frac{2}{3} 	ext{ sq. units}$.

The area (in $sq. \text{ units}$) bounded by the curve $y=x|x|$,$X$-axis and the lines $x=-1$ and $x=1$ is

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