The area (in square units) in the first quadrant bounded by the curve $y=x^2+2$ and the lines $y=x+1, x=0$ and $x=3$, is

Question

Aicharya · Accepted Answer

$\begin{aligned} 	ext { Required area } & =\int_0^3\left(x^2+2ight)-(x+1) \mathrm{d} x \ & =\int_0^3\left(x^2-x+1ight) \mathrm{d} x \ & =\left[\frac{x^3}{3}-\frac{x^2}{2}+xight]_0^3=\frac{15}{2} 	ext { sq. units }\end{aligned}$

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

Solution