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NDA 2021 Maths · Mathematics · Application of Integrals · medium

The area under the curve y=x√(1+x^2 between x=0 and x=1 is

Single correct option

The area under the curve y=x1+x2y=x\sqrt{1+x^2} between x=0x=0 and x=1x=1 is
  1. A. 223\dfrac{2\sqrt2}{3} square units
  2. B. 2−13\dfrac{\sqrt2-1}{3} square units
  3. C. 22−13\dfrac{2\sqrt2-1}{3} square units
  4. D. 22−12\dfrac{2\sqrt2-1}{2} square units
Show answer and solution

Answer: (C) 22−13\dfrac{2\sqrt2-1}{3} square units

Put u=1+x2u=1+x^2, du=2x dxdu=2x\,dx: area =12∫12u1/2du=12⋅23(22−1)=22−13=\tfrac12\int_1^2u^{1/2}du=\tfrac12\cdot\tfrac23\left(2\sqrt2-1\right)=\dfrac{2\sqrt2-1}{3}. The lower limit contributes −1/3-1/3, which is easy to drop.
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Original question written in the NDA 2021 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake