The area of the region bounded by the curves | vedclass

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(C) Given curves are $x=y^2-2$ and $x=y$.To find the points of intersection, we substitute $x=y$ into $x=y^2-2$:$y = y^2 - 2$$y^2 - y - 2 = 0$$(y-2)(y

Vedclass · Accepted Answer

$\frac{9}{2}$

Vedclass · Answer

(C) Given curves are $x=y^2-2$ and $x=y$.To find the points of intersection, we substitute $x=y$ into $x=y^2-2$:$y = y^2 - 2$$y^2 - y - 2 = 0$$(y-2)(y+1) = 0$So, $y=2$ and $y=-1$.For $y=2$, $x=2$. For $y=-1$, $x=-1$.The points of intersection are $(-1, -1)$ and $(2, 2)$.The area $A$ is given by the integral of the right curve minus the left curve with respect to $y$ from $y=-1$ to $y=2$:$A = \int_{-1}^{2} (y - (y^2 - 2)) \, dy$$A = \int_{-1}^{2} (y - y^2 + 2) \, dy$$A = \left[ \frac{y^2}{2} - \frac{y^3}{3} + 2y \right]_{-1}^{2}$$A = \left( \frac{2^2}{2} - \frac{2^3}{3} + 2(2) \right) - \left( \frac{(-1)^2}{2} - \frac{(-1)^3}{3} + 2(-1) \right)$$A = \left( 2 - \frac{8}{3} + 4 \right) - \left( \frac{1}{2} + \frac{1}{3} - 2 \right)$$A = \left( 6 - \frac{8}{3} \right) - \left( \frac{3+2-12}{6} \right)$$A = \frac{10}{3} - \left( -\frac{7}{6} \right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$ square units.

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

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