SATHEE: Definite Integration Question 25

Definite Integration Question 25

Question 25

The integral $\int_{- 1 / 2}^{1 / 2} [x] + \log \frac{1 + x}{1 - x} d x$ equals(2002, 1M) (a) $- \frac{1}{2}$ (b) 0 (c) 1 (d) $\log \frac{1}{2}$

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Solution:

$\int_{- 1 / 2}^{1 / 2} [x] + \log \frac{1 + x}{1 - x} d x$

$= \int_{- 1 / 2}^{1 / 2} [x] d x + \int_{- 1 / 2}^{1 / 2} \log \frac{1 + x}{1 - x} d x$

$= \int_{- 1 / 2}^{1 / 2} [x] d x + 0 ∵ \log \frac{1 + x}{1 - x}$ is an odd function

$= \int_{- 1 / 2}^{0} [x] d x + \int_{0}^{1 / 2} [x] d x = \int_{- 1 / 2}^{0} (- 1) d x + \int_{0}^{1 / 2} (0) d x$

$= - [x]_{- 1 / 2}^{0} = - 0 + \frac{1}{2} = - \frac{1}{2}$

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Definite Integration Question 25

Question 25

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