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CBSE Class 12 · 2011 · Definite IntegralsLong Answer

Evaluate:

\int_{0}^{1} lo g (\frac{1}{x} - 1) d x

Answer: $0$

Solution

Let

I = \int_{0}^{1} lo g (\frac{1}{x} - 1) d x = \int_{0}^{1} lo g \frac{1 - x}{x} d x

.
Apply

x \to 1 - x

:

I = \int_{0}^{1} lo g \frac{x}{1 - x} d x = - \int_{0}^{1} lo g \frac{1 - x}{x} d x = - I

.
Therefore

2 I = 0

, i.e.

I = 0

.

CBSE Class 12 Board · 2011 · Subjective

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