There are (in square units) of the region bounded by the parabola $y=x^2+2$ and the lines $y=x+1, x=0$ and $x=3$ is

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Aicharya · Accepted Answer

Required area $\begin{aligned} & =\int_0^3\left\{\left(x^2+2ight)-(x+1)ight\} d x \ & =\left[\frac{x^3}{3}+2 x-\frac{x^2}{2}-xight]_0^3 \ & =\left(9+6-\frac{9}{2}-3ight)-0=\frac{15}{2} \end{aligned}$

There are (in square units) of the region bounded by the parabola $y=x^2+2$ and the lines $y=x+1, x=0$ and…

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