Question

The area $(insq.units)$ of the region described by $\{(x,y)y^2\le 2xandy\ge 4x-1\}$ is

• A. $\frac{7}{32}$
• B. $\frac{5}{64}$
• C. $\frac{15}{64}$
• D. $\frac{9}{32}$

Answer 1

The correct option is D $\frac{9}{32}$

$x=0.5$

$x=0.15$

$y^2⩽2x$

$y⩾4x-1$

${\int }_{0.125}^{0.5}{\int }_{4x-1}^{\sqrt{2x}}1.dydx+{\int }_0^{0.125}{\int }_{-\sqrt{2x}}^{\sqrt{2x}}1.dy.dx$

$={\int }_{0.5}^{0.125}\left(\sqrt{2x-4x+1}\right)dx+{\int }_0^{0.125}2\sqrt{2x}dx$

$={(\sqrt{2}\frac{x^{3/2}}{3/2}-\frac{4x^2}{2}+x)}_{0.125}^{0.5}+2\sqrt{2}{\left(\frac{x^{3/2}}{3/2}\right)}_0^{0.125}$

$=(\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-\frac{1}{24}+\frac{1}{32}-\frac{1}{8})+(\frac{1}{12}-0)$

$=\frac{9}{32}$