Q42 / 90 · JEE Main 2012 (26 May Online) · Mathematics

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

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

Solution

Intersecting points are $x=1,4$ $$ \begin{aligned} & \therefore \text { Required area }=\int_1^4\left[2 \sqrt{x}-\left(\frac{2 x+4}{3}\right)\right] d x \\ & =\left.\frac{2 x^{3 / 2}}{3 / 2}\right|_1 ^4-\left.\frac{2 x^2}{3 \times 2}\right|_1 ^4-\left.\frac{4}{3} x\right|_1 ^4 \end{aligned} $$ $$ \begin{aligned} & =\frac{4}{3}\left(4^{3 / 2}-1^{3 / 2}\right)-\frac{1}{3}(16-1)\left[\frac{4}{3}(4)-\frac{4}{3}\right] \\ & =\frac{4}{3}(7)-5-4=\frac{28}{3}-9=\frac{28-27}{3}=\frac{1}{3} \end{aligned} $$

Hence, the correct answer is option: 'B'.

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