Question

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

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

Answer 1

Answer: (D)

$\frac{9}{32}$

Solving the curve $y^2=2x$ and $y=4x-1$ we get the point of intersection
of the curves which are $(\frac{1}{2},1)$ and $(\frac{1}{8},-\frac{1}{2})$

Hence required area is,
$={\int }_{-\frac{1}{2}}^1(\frac{y+1}{4}-\frac{1}{2}{y}^2)dy$
$=\frac{1}{4}{\int }_{-\frac{1}{2}}^1(y+1-2y^2)dy=\frac{1}{4}{(\frac{y^2}{2}+y-\frac{2}{3}{y}^3)}_{-\frac{1}{2}}^1=\frac{9}{32}$
Note: By plotting the curves on graph it could be understood very easily.