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)

$\frac{1}{3}$

Answer 2

Intersecting points are x = 1, 4 $\therefore$ Required area $= \int\limits^{4}_{1} \left[2 \sqrt{x} - \left(\frac{2x+4}{3}\right)\right]dx$ $= \frac{2x ^{\frac{3}{2}}}{\frac{3}{2}} \Bigg|^{4}_{1} - \frac{2x^{2}}{3 \times2} \Bigg|^{4}_{1} - \frac{4}{3} x \bigg|^{4}_{1}$ $= \frac{4}{3}\left(4^{\frac{3}{2}} - 1^{\frac{3}{2}}\right) - \frac{1}{3} \left(16-1\right) - \left[\frac{4}{3} \left(4\right) - \frac{4}{3}\right]$ $= \frac{4}{3}\left(7\right) - 5-4 = \frac{28}{3} - 9 = \frac{28-27}{3} = \frac{1}{3}$