Question

The area bounded by the parabola $y^2=4x$ and the line $2x-3y+4=0$, in square unit, is

• A. $\frac{2}{5}$
• B. $\frac{1}{3}$
• C. 1
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

Intersecting points are x = 1, 4
$\therefore$ Required area $=\underset{1}{\overset{4}{\int }}\left[2\sqrt{x}-\left(\frac{2x+4}{3}\right)\right]dx$
$=\frac{2x^{\frac{3}{2}}}{\frac{3}{2}}{|}_1^4-\frac{2x^2}{3\times 2}{|}_1^4-\frac{4}{3}x{|}_1^4$
$=\frac{4}{3}\left(4^{\frac{3}{2}}-1^{\frac{3}{2}}\right)-\frac{1}{3}\left(16-1\right)-\left[\frac{4}{3}\left(4\right)-\frac{4}{3}\right]$
$=\frac{4}{3}\left(7\right)-5-4=\frac{28}{3}-9=\frac{28-27}{3}=\frac{1}{3}$