The area (in sq. units), in the first quadrant bounded by the curve $y=x^2+2$ and the lines $y=x+1, x=0$ and $x=2$, is

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Aicharya · Accepted Answer

$\begin{aligned} 	ext { Required area } & =\int_0^2\left[\left(x^2+2ight)-(x+1)ight] \mathrm{d} x \ & =\int_0^2\left(x^2-x+1ight) \mathrm{d} x \ & =\left[\frac{x^3}{3}-\frac{x^2}{2}+xight]_0^2 \ & =\frac{8}{3}-2+2-0 \ & =\frac{8}{3} 	ext { sq. units }\end{aligned}$

The area (in sq. units), in the first quadrant bounded by the curve $y=x^2+2$ and the lines $y=x+1, x=0$ and…

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