Advertisements
Advertisements
Question
Advertisements
Solution
\[Let\ I = \int_0^1 \log\left( \frac{1}{x} - 1 \right) d x ...............(1)\]
\[ = \int_0^1 \log\left( \frac{1}{1 - x} - 1 \right) d x ...............\left[\text{Using }\int_0^a f(x) dx = \int_0^a f(a - x) dx \right]\]
\[ I = \int_0^1 \log\left( \frac{x}{1 - x} \right) dx ...............(2)\]
\[\text{Adding (1) and (2)}\]
\[2I = \int_0^1 \log\left( \frac{1 - x}{x} \right) + \log\left( \frac{x}{1 - x} \right) dx\]
\[ = \int_0^1 \log\left( \frac{1 - x}{x} \times \frac{x}{1 - x} \right) dx\]
\[ = \int_0^1 \log1 dx \]
\[ = 0\]
\[Hence\ I = 0\]
APPEARS IN
RELATED QUESTIONS
Evaluate each of the following integral:
\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]
Evaluate the following integral:
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
`int_0^(2a)f(x)dx`
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
Using second fundamental theorem, evaluate the following:
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x`
Evaluate the following:
f(x) = `{{:("c"x",", 0 < x < 1),(0",", "otherwise"):}` Find 'c" if `int_0^1 "f"(x) "d"x` = 2
Evaluate the following:
`Γ (9/2)`
Choose the correct alternative:
Γ(1) is
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
