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NDA 2020 Maths · Mathematics · Integrals · medium

The integral ∫0^1x^2ln x dx converges at the lower limit, and its value is

Single correct option

The integral ∫01x2ln⁡x dx\displaystyle\int_0^1x^2\ln x\,dx converges at the lower limit, and its value is
  1. A. 19\dfrac19
  2. B. −13-\dfrac13
  3. C. −29-\dfrac29
  4. D. −19-\dfrac19
Show answer and solution

Answer: (D) −19-\dfrac19

By parts with u=ln⁡xu=\ln x and dv=x2dxdv=x^2dx: ∫x2ln⁡x dx=x33ln⁡x−x39\displaystyle\int x^2\ln x\,dx=\dfrac{x^3}{3}\ln x-\dfrac{x^3}{9}. At x=1x=1 this is −19-\dfrac19, and as x→0+x\to0^+ the term x3ln⁡x→0x^3\ln x\to0, so the lower limit contributes nothing. The value is −19-\dfrac19, negative because ln⁡x<0\ln x<0 on (0,1)(0,1), which rules out 19\dfrac19. The option −13-\dfrac13 drops the factor 13\dfrac13 in the second integral.
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Original question written in the NDA 2020 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