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Question

(A) The given curves are $y=x^2$ and $y=x^3$. To find the intersection points, set $x^2 = x^3$, which implies $x^2(1-x) = 0$. Thus, the curves interse

Vedclass · Accepted Answer

$\frac{1}{12}$

Vedclass · Answer

(A) The given curves are $y=x^2$ and $y=x^3$. To find the intersection points, set $x^2 = x^3$, which implies $x^2(1-x) = 0$. Thus, the curves intersect at $x=0$ and $x=1$. In the interval $[0, 1]$, $x^2 \ge x^3$. The area $A$ is given by the integral: $A = \int_0^1 (x^2 - x^3) dx$ $A = \left[ \frac{x^3}{3} - \frac{x^4}{4} ight]_0^1$ $A = \left( \frac{1}{3} - \frac{1}{4} ight) - (0 - 0)$ $A = \frac{4-3}{12} = \frac{1}{12} 	ext{ square units}$.

The area (in square unit) of the region enclosed by the curves $y=x^2$ and $y=x^3$ is

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