Question

The area (in sq. units) of the region described by $\left\{(x, y): y^{2} \leq 2 x\right.$ and $\left.y \geq 4 x-1\right\}$ is

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

Answer 1

Answer: (D)

$y^{2}=2 x$
$\frac{y^{2}}{2}=\frac{y+1}{4}$
$2 y^{2}-y-1=0$
$2 y^{2}-2 y+y-1=0$
$(2 y+1)(y-1)$
$A=\int\limits_{\frac{-1}{2}}^{1}\left(\frac{y+1}{4}-\left(\frac{y^{2}}{2}\right)\right) d$
$A=\left(\frac{\frac{y^{2}}{2}+y}{4}\right)_{-\frac{1}{2}}^{1}-\left(\frac{y^{3}}{6}\right)_{-\frac{1}{2}}^{1}$
$A=\left(\frac{y^{2}+2 y}{8}\right)_{-\frac{1}{2}}^{1}-\left(\frac{y^{3}}{6}\right)_{-\frac{1}{2}}^{1}$
$A=\left(\frac{3}{8}-\frac{\left\{\frac{1}{4}-1\right\}}{8}\right)-\left(\frac{1}{6}+\frac{1}{48}\right)$
$A=\left(\frac{3}{8}+\frac{3}{32}\right)-\left(\frac{8+1}{48}\right)=\frac{12+3}{32}-\frac{9}{48}$
$A=\frac{15}{32}-\frac{9}{48}=\frac{3}{16}\left(\frac{5}{2}-\frac{3}{3}\right)$
$=\frac{3}{16} \times\left(\frac{15-6}{6}\right)=\frac{3 \times 9}{16 \times 6}=\frac{9}{32}$