The area bounded by $y=x^2+2, x$-axis, $x=1$ and $x=2$ is
- $\frac{16}{3}$ sq unit
- $\frac{17}{3}$ sq unit
- $\frac{13}{3}$ sq unit
- $\frac{20}{3}$ sq unit
Solution

$\begin{aligned} & \text { Required area }=\text { Area of curve } A B C D \\ & \quad=\int_1^2 y d x=\int_1^2\left(x^2+2\right) d x=\left[\frac{x^3}{3}+2 x\right]_1^2 \\ & \quad=\left(\frac{8}{3}+4\right)-\left(\frac{1}{3}+2\right)=\frac{20}{3}-\frac{7}{3}=\frac{13}{3} \text { sq unit }\end{aligned}$
Asked in: AP EAMCET 2004