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NDA 2021 Maths · Mathematics · Definite Integrals · easy

∫0^πsin x dx=2. 2. ∫0^2πsin x dx=2.

Single correct option

Which of the statements is/are correct? 1. ∫0πsin⁡x dx=2\displaystyle\int_0^{\pi}\sin x\,dx=2. 2. ∫02πsin⁡x dx=2\displaystyle\int_0^{2\pi}\sin x\,dx=2.
  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: (A) 1 only

∫0πsin⁡x dx=[−cos⁡x]0π=2\int_0^\pi\sin x\,dx=[-\cos x]_0^\pi=2, so statement 1 is correct. Over a full period the positive and negative arches cancel: ∫02πsin⁡x dx=0\int_0^{2\pi}\sin x\,dx=0, so statement 2 is wrong.
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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