The area of the region, bounded by the parabola $y=x^2+2$ and the lines $y=x, x=0$ and $x=3$, is

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Aicharya · Accepted Answer

$\begin{aligned} 	ext { Required area } & =\int_0^3\left(x^2+2-\dot{x}ight) \mathrm{d} x \ & =\left[\frac{x^3}{3}+2 x-\frac{x^2}{2}ight]_0^3 \ & =9+6-\frac{9}{2}-0 \ & =\frac{21}{2} 	ext { sq. unit }\end{aligned}$

The area of the region, bounded by the parabola $y=x^2+2$ and the lines $y=x, x=0$ and $x=3$, is

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