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JEE Main 2019 · Mathematics · Area Under Curves · medium

Find the area enclosed between the curve y^2=4x and the straight line y=x. It is

Single correct option

Find the area enclosed between the curve y2=4xy^2=4x and the straight line y=xy=x. It is
  1. A. 163\frac{16}{3}
  2. B. 43\frac43
  3. C. 8
  4. D. 83\frac83
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Answer: (D) 83\frac83

Intersection: x2=4x⇒x=0,4x^2=4x\Rightarrow x=0,4. Between them y=2xy=2\sqrt x lies above y=xy=x, so area =∫04(2x−x) dx=[43x3/2−x22]04=323−8=83=\int_0^4(2\sqrt x-x)\,dx=\left[\tfrac43x^{3/2}-\tfrac{x^2}{2}\right]_0^4=\tfrac{32}{3}-8=\tfrac83.
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Original question written in the JEE Main 2019 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake