Question

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

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

Answer 1

Answer: (C)

$0 \leq y \leq x^{2}+1,0 \leq y \leq x+1, \frac{1}{2} \leq x \leq 2$

Required area $=\int_{1 / 2}^{1}\left(x^{2}+1\right) d x+\frac{1}{2}(2+3) \times 1$ $
=\frac{19}{24}+\frac{5}{2}=\frac{79}{24} $