Question:
If the area (in sq. units) of the region $\left\{(x, y): y^{2} \leq 4 x, x+y \leq 1, x \geq 0, y \geq O\right\}$ is $a \sqrt{2}+b$, then $a-b$ is equal to :
Correct Option: , 3
Solution:
$\left\{(\mathrm{x}, \mathrm{y}): \mathrm{y}^{2} \leq 4 \mathrm{x}, \mathrm{x}+\mathrm{y} \leq 1, \mathrm{x} \geq 0, \mathrm{y} \geq 0\right\}$
$\mathrm{A} \int_{0}^{3-2 \sqrt{2}} 2 \sqrt{\mathrm{x}} \mathrm{dx}+\frac{1}{2}(1-(3-2 \sqrt{2}))(1-(3-2 \sqrt{2}))$
$=\frac{2\left[x^{3 / 2}\right]_{0}^{3-2 \sqrt{2}}}{3 / 2}+\frac{1}{2}(2 \sqrt{2}-2)(2 \sqrt{2}-2)$
$=\frac{8 \sqrt{2}}{3}+\left(-\frac{10}{3}\right) \quad$ Activate Windows
$\mathrm{a}=\frac{8}{3}, \mathrm{~b}=-\frac{10}{3}$
$a-b=6$