Question

The area (in sq. units) of the region $\{(x,y):0\le y\le x^2+1,0\le y\le x+1,\frac{1}{2}\le x\le 2\}$ is

• A. $\frac{23}{16}$
• B. $\frac{79}{16}$
• C. $\frac{23}{6}$
• D. $\frac{79}{24}$

Answer 1

The correct option is D $\frac{79}{24}$


$A=\text{Area}\left(ABCD\right)+\text{Area}\left(CDEF\right)+\text{Area}\left(CFG\right)$
$A=\underset{1/2}{\overset{1}{\int }}(x^2+1)dx+1\times 2+\frac{1}{2}\times 1\times 1$
$\Rightarrow A={(\frac{x^3}{3}+x)}_{1/2}^1+2+\frac{1}{2}$
$\Rightarrow A=(\frac{4}{3}-\frac{13}{24})+\frac{5}{2}$
$\Rightarrow A=\frac{19}{24}+\frac{5}{2}$
$\therefore A=\frac{79}{24}$

Alternate Solution:
$A=\underset{1/2}{\overset{1}{\int }}(x^2+1)dx+\underset{1}{\overset{2}{\int }}(x+1)dx$
$\Rightarrow A={(\frac{x^3}{3}+x)}_{1/2}^1+{(\frac{x^2}{2}+x)}_1^2$
$\Rightarrow A=\frac{19}{24}+\frac{5}{2}$
$\therefore A=\frac{79}{24}$