Question

The area in square units of the region bounded by $y^2=9x$ and $y=3x$ is

• A. 2
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. 1

Answer 1

Answer: (C)

Intersecting points of $y^2=9x$ and $y=3x$ is $(0,0)$ and $(1,3)$.
$\therefore$ Required Area = $\underset{0}{\overset{1}{\int }}(\sqrt{9x}-3x)dx$
$=3\underset{0}{\overset{1}{\int }}\left(\sqrt{x}-x\right)dx=3{\left[\frac{x^{3/2}}{3/2}-\frac{x^2}{2}\right]}_0^1$
$=3\left(\frac{2}{3}-\frac{1}{2}\right)=3\left(\frac{1}{6}\right)=\frac{1}{2}$ sq. unit