Question

The area of the region $\left\{\left(x,y\right):0\le y\le x^2+1,0\le y\le x+1,0\le x\le 2\right\}$ is

• A. $\frac{11}{6}$
• B. $\frac{15}{4}$
• C. $\frac{25}{6}$
• D. $\frac{23}{6}$

Answer 1

Answer: (D)

$\frac{23}{6}$

Area $={\int }_0^1(x^2+1)dx+{\int }_1^2(x+1)dx$

$={[\frac{x^3}{3}+x]}_0^1+{[\frac{x^2}{2}+x]}_1^2$

$=[\frac{1}{3}+1]+[\frac{2^2}{2}+2]-[\frac{1^2}{2}+1]$

$=\frac{4}{3}+4-\frac{1}{2}-7$

$=3+\frac{5}{6}$

$=\frac{23}{6}$