First slide

Evaluation of definite integrals

Question

If $I_{1} = \int_{x}^{1} \frac{d t}{1 + t^{2}} and I_{2} = \int_{1}^{1 / x} \frac{d t}{1 + t^{2}} for x > 0$ then

Moderate

A

$I_{1} > I_{2}$
B

$I_{1} = I_{2}$
C

$I_{2} > I_{1}$
D

$I_{2} = (π / 2) - \tan^{- 1} x$

Solution

In $I_{2}, put t = 1 / u$ to obtain
$I_{2} = \int_{1}^{x} \frac{(- 1 / u^{2}) d u}{1 + 1 / u^{2}} = \int_{x}^{1} \frac{d t}{1 + t^{2}} = I_{1}$

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