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NDA 2025 Maths · Mathematics · Definite Integrals · hard

The value of ∫0^2x√(4-x^2 dx is

Single correct option

The value of ∫02x4−x2 dx\displaystyle\int_0^2x\sqrt{4-x^2}\,dx is
  1. A. 163\dfrac{16}3
  2. B. 43\dfrac43
  3. C. 83\dfrac83
  4. D. 22
Show answer and solution

Answer: (C) 83\dfrac83

Put u=4−x2u=4-x^2, du=−2x dxdu=-2x\,dx. Then ∫02x4−x2dx=12∫04u du=12⋅23⋅8=83\int_0^2x\sqrt{4-x^2}dx=\frac12\int_0^4\sqrt u\,du=\frac12\cdot\frac23\cdot8=\frac83. The value 163\frac{16}3 forgets the factor 12\frac12, and 43\frac43 halves twice. Changing the limits to u=4u=4 and u=0u=0 avoids back-substitution.
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Original question written in the NDA 2025 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