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(A) The area of the region enclosed by the parabola $x^{2}=y,$ the line $y=x+2,$ and the $x-$ axis is the shaded region shown in the figure.The line $

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(A) The area of the region enclosed by the parabola $x^{2}=y,$ the line $y=x+2,$ and the $x-$ axis is the shaded region shown in the figure.The line $y=x+2$ intersects the $x-$ axis at $x=-2$ (where $y=0$).The parabola $x^{2}=y$ and the line $y=x+2$ intersect where $x^{2}=x+2.$$x^{2}-x-2=0 \Rightarrow (x-2)(x+1)=0.$Since the region is in the second quadrant,we take the intersection point $x=-1.$The required area is the sum of the area under the line from $x=-2$ to $x=-1$ and the area under the parabola from $x=-1$ to $x=0$ (or simply the area under the line from $x=-2$ to $x=-1$ plus the area under the parabola from $x=-1$ to $x=0$ is not correct; looking at the figure,it is the area under the line from $x=-2$ to $x=-1$ plus the area under the parabola from $x=-1$ to $x=0$ is actually the area bounded by the line and the $x-$ axis from $x=-2$ to $x=-1$ plus the area bounded by the parabola and the $x-$ axis from $x=-1$ to $x=0$).Area $= \int_{-2}^{-1} (x+2) dx + \int_{-1}^{0} x^{2} dx$$= \left[ \frac{x^{2}}{2} + 2x ight]_{-2}^{-1} + \left[ \frac{x^{3}}{3} ight]_{-1}^{0}$$= \left( (\frac{1}{2} - 2) - (\frac{4}{2} - 4) ight) + (0 - (-\frac{1}{3}))$$= (-\frac{3}{2} - (-2)) + \frac{1}{3}$$= \frac{1}{2} + \frac{1}{3} = \frac{5}{6} 	ext{ sq. units.}$

Find the area of the region enclosed by the parabola $x^{2}=y,$ the line $y=x+2$ and the $x-$ axis.

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