The area bounded by the curves y^2 - x = 0 an | vedclass

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(B) The given curves are $y^2 = x$ and $y = x^2$. To find the points of intersection,substitute $x = y^2$ into $y = x^2$,which gives $y = (y^2)^2 = y^

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$\frac{1}{3}$

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(B) The given curves are $y^2 = x$ and $y = x^2$. To find the points of intersection,substitute $x = y^2$ into $y = x^2$,which gives $y = (y^2)^2 = y^4$. This implies $y^4 - y = 0$,so $y(y^3 - 1) = 0$. The intersection points are at $y = 0$ and $y = 1$. For $y = 0$,$x = 0$,and for $y = 1$,$x = 1$. The area $A$ bounded by the curves is given by the integral: $A = \int_0^1 (\sqrt{x} - x^2) dx$ $A = \left[ \frac{x^{3/2}}{3/2} - \frac{x^3}{3} \right]_0^1$ $A = \left( \frac{2}{3}(1)^{3/2} - \frac{1}{3}(1)^3 \right) - (0 - 0)$ $A = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}$ Thus,the area is $\frac{1}{3}$ square units.

The area bounded by the curves $y^2 - x = 0$ and $y - x^2 = 0$ is

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