Advertisements
Advertisements
Question
Solution
\[Let\ I = \int_{- 1}^1 \log\frac{2 - x}{2 + x} d x\]
\[Here\ f\left( x \right) = \log\left( \frac{2 - x}{2 + x} \right)\]
\[f\left( - x \right) = \log\left( \frac{2 + x}{2 - x} \right)\]
\[ = - \log\left( \frac{2 - x}{2 + x} \right)\]
\[ = - f\left( x \right)\]
\[\text{Hence} f\left( x \right) \text{is an odd function}, \]
Therefore,
\[I = 0\]
APPEARS IN
RELATED QUESTIONS
Evaluate each of the following integral:
\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = ______.
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
