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Question: Answered & Verified by Expert
If $\mathrm{I}=\int_0^{\mathrm{I}} \frac{\mathrm{dx}}{1+\mathrm{x}^{\pi / 2}}$, then
MathematicsDefinite IntegrationWBJEEWBJEE 2010
Options:
  • A $\log _{\mathrm{e}} 2 < 1 < \pi / 4$
  • B $\log _{\mathrm{e}} 2>1$
  • C $\mathrm{I}=\pi / 4$
  • D $\mathrm{I}=\log _{\mathrm{e}} 2$
Solution:
1100 Upvotes Verified Answer
The correct answer is: $\log _{\mathrm{e}} 2 < 1 < \pi / 4$
Hints: $\mathrm{x}^2 < \mathrm{x}^{\frac{\pi}{2}} < \mathrm{x}, \quad 1+\mathrm{x}^2 < 1+\mathrm{x}^{\frac{\pi}{2}} < 1+\mathrm{x}$
$\frac{1}{1+x^2}>\frac{1}{1+x^{\frac{\pi}{2}}}>\frac{1}{1+x}$
$\frac{\pi}{4}>\mathrm{I}>(\log (1+\mathrm{x})), \quad \frac{\pi}{4}>\mathrm{I}>\log 2$

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