Definite Integration Question 25
Question 25
- The integral $\int_{-1 / 2}^{1 / 2}[x]+\log \frac{1+x}{1-x} \quad d x$ equals(2002, 1M) (a) $-\frac{1}{2}$ (b) 0 (c) 1 (d) $\log \frac{1}{2}$
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Solution:
- $\int_{-1 / 2}^{1 / 2}[x]+\log \frac{1+x}{1-x} d x$
$=\int_{-1 / 2}^{1 / 2}[x] d x+\int_{-1 / 2}^{1 / 2} \log \frac{1+x}{1-x} d x$
$=\int_{-1 / 2}^{1 / 2}[x] d x+0 \quad \because \log \frac{1+x}{1-x}$ is an odd function
$=\int_{-1 / 2}^{0}[x] d x+\int_{0}^{1 / 2}[x] d x=\int_{-1 / 2}^{0}(-1) d x+\int_{0}^{1 / 2}(0) d x$
$=-[x]_{-1 / 2}^{0}=-0+\frac{1}{2}=-\frac{1}{2}$
