If the area (in sq. units) of the region {(x, y) : y2 4x, x + y 1, x 0, y 0} is , then a

Engineering

Mathematics

Area between Two Curves

Question

If the area (in sq. units) of the region {(x, y) : y² $\geq$ 4x, x + y $\leq$ 1, x $\geq$ 0, y $\geq$ 0} is $a \sqrt{2} + b$ , then a – b is equal to

$\frac{10}{3}$

$\frac{8}{3}$

$- \frac{2}{3}$

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{(x, y)} : y² $\geq$ 4x, x + y $\leq$ 1, x $\geq$ 0, y $\geq$ 0}

$A \int_{0}^{3 - 2 \sqrt{2}} 2 \sqrt{x} dx + \frac{1}{2} (1 - (3 - 2 \sqrt{2})) (1 - (3 - 2 \sqrt{2}))$

$= \frac{2 {[x^{3 / 2}]}_{0}^{3 - 2 \sqrt{2}}}{3 / 2} + \frac{1}{2} (2 \sqrt{2} - 2) (2 \sqrt{2} - 2)$

$= \frac{8 \sqrt{2}}{3} + (- \frac{10}{3})$

$a = \frac{8}{3}, b = - \frac{10}{3}$

a – b = 6