Question

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

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

Answer 1

Answer: (D)

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