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(D) To find the area of the region bounded by $y=x$ and $y=x^3$,we first find the intersection points by setting $x^3 = x$.This gives $x^3 - x = 0$,so

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$0.5 	ext{ sq unit}$

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(D) To find the area of the region bounded by $y=x$ and $y=x^3$,we first find the intersection points by setting $x^3 = x$.This gives $x^3 - x = 0$,so $x(x^2 - 1) = 0$,which implies $x = -1, 0, 1$.The area is symmetric about the origin.Area $= 2 \int_0^1 (x - x^3) dx$$= 2 \left[ \frac{x^2}{2} - \frac{x^4}{4} ight]_0^1$$= 2 \left( \frac{1}{2} - \frac{1}{4} ight)$$= 2 \left( \frac{1}{4} ight) = 0.5 	ext{ sq units}$.

The area of the region bounded by the line $y=x$ and the curve $y=x^3$ is

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