The area (in sq. units) bounded by x^2=y,y=x+ | vedclass

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(C) The region is bounded by the parabola $x^2=y$,the line $y=x+2$,and the $X$-axis.The intersection points of the parabola $y=x^2$ and the line $y=x+

Vedclass · Accepted Answer

$\frac{5}{6}$

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(C) The region is bounded by the parabola $x^2=y$,the line $y=x+2$,and the $X$-axis.The intersection points of the parabola $y=x^2$ and the line $y=x+2$ are found by setting $x^2=x+2$,which gives $x^2-x-2=0$,so $(x-2)(x+1)=0$. Thus,$x=-1$ and $x=2$.The line $y=x+2$ intersects the $X$-axis at $x=-2$ (where $y=0$).The required area is the region bounded by the line from $x=-2$ to $x=-1$ and the parabola from $x=-1$ to $x=0$ (since the region is bounded by the $X$-axis,we consider the area under the curve down to $y=0$).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) + \left( 0 - (-\frac{1}{3}) ight)$$= (-\frac{3}{2} - (-2)) + \frac{1}{3}$$= \frac{1}{2} + \frac{1}{3} = \frac{5}{6} 	ext{ sq. units}$.

The area (in sq. units) bounded by $x^2=y$,$y=x+2$ and the $X$-axis is

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