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 :
Correct Option: , 3
Solution:
$0 \leq y \leq x^{2}+1,0 \leq y \leq x+1, \frac{1}{2} \leq x \leq 2$
Required area $=\int_{1 / 2}^{1}\left(x^{2}+1\right) d x+\frac{1}{2}(2+3) \times 1$
$=\frac{19}{24}+\frac{5}{2}=\frac{79}{24}$
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