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BITSAT 2009
Exam
BITSAT
Year
2009
Subject
Mathematics
Chapter
Definite Integration

BITSAT 2009 · Mathematics · Definite Integration

Evaluate : (∫_0^1 dx 2-x^2 ).

Evaluate : (\int_0^1 dx2−x2\frac{\mathrm{dx}}{\sqrt{2-\mathrm{x}^2}}).

Options

  1. A(\pi / 4)
  2. B(\pi)
  3. C(\pi / 2)
  4. D(\pi / 3)

Correct answer

A. (\pi / 4)

Step-by-step solution

∫dx2−x2=sin⁡−1x2+cSo ∫01dx2−x2=sin⁡−1x2∣01=sin⁡−1(12)+c−sin⁡−1(0)−c=\ppi4−0=π4\begin{aligned} & \int \frac{\mathrm{dx}}{\sqrt{2-\mathrm{x}^2}}=\sin ^{-1} \frac{\mathrm{x}}{\sqrt{2}}+\mathrm{c}\\ & \text{So } \int_0^1 \frac{\mathrm{dx}}{\sqrt{2-\mathrm{x}^2}}=\left.\sin ^{-1} \frac{\mathrm{x}}{\sqrt{2}}\right|_0 ^1 \\ = & \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)+c-\sin ^{-1}(0)-c=\frac{\ppi}{4}-0=\frac{\pi}{4} \end{aligned})$
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