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NDA 2020 Maths · Mathematics · Definite Integrals · easy

Two statements about definite integrals over the interval [0,(π)/(2)] are given below. 1. ∫0^π/2sin x dx=∫0^π/2cos x dx. 2. ∫0^π/2sin ^2x…

Single correct option

Two statements about definite integrals over the interval [0,π2]\left[0,\dfrac\pi2\right] are given below.
1. ∫0π/2sin⁡x dx=∫0π/2cos⁡x dx\displaystyle\int_0^{\pi/2}\sin x\,dx=\int_0^{\pi/2}\cos x\,dx.
2. ∫0π/2sin⁡2x dx=∫0π/2cos⁡2x dx\displaystyle\int_0^{\pi/2}\sin^2x\,dx=\int_0^{\pi/2}\cos^2x\,dx.
Which of the statements is/are correct?
  1. A. 1 only
  2. B. 2 only
  3. C. Both 1 and 2
  4. D. Neither 1 nor 2
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Answer: (C) Both 1 and 2

Replacing xx by π2−x\dfrac\pi2-x turns sin⁡x\sin x into cos⁡x\cos x inside a definite integral over [0,π2]\left[0,\dfrac\pi2\right], because ∫abf(x)dx=∫abf(a+b−x)dx\displaystyle\int_a^bf(x)dx=\int_a^bf(a+b-x)dx. Hence both integrals in statement 1 equal 11, and both in statement 2 equal π4\dfrac\pi4. The functions look different, but the symmetric interval makes the areas equal. So both statements are correct; no calculation of the antiderivatives is needed.
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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