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Area of the region bounded by the curve $y=x^2+2$ and the lines $y=x, x=0$ and $x=3$ is
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$\frac{21}{2} \mathrm{sq} \cdot$ units
$\begin{aligned} & \text { Required area }=\int_0^3\left(x^2+2-x\right) d x=\left[\frac{x^3}{3}+2 x-\frac{x^2}{2}\right]_0^3 \\ & =\frac{27}{3}+6-\frac{9}{2}=\frac{21}{2}\end{aligned}$
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