Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_{- 1}^1 \log\frac{2 - x}{2 + x} d x\]
\[Here\ f\left( x \right) = \log\left( \frac{2 - x}{2 + x} \right)\]
\[f\left( - x \right) = \log\left( \frac{2 + x}{2 - x} \right)\]
\[ = - \log\left( \frac{2 - x}{2 + x} \right)\]
\[ = - f\left( x \right)\]
\[\text{Hence} f\left( x \right) \text{is an odd function}, \]
Therefore,
\[I = 0\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\] then the value of I10 + 90I8 is
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
\[\int\limits_0^4 x dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
Evaluate the following using properties of definite integral:
`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x) "d"x`
Choose the correct alternative:
If n > 0, then Γ(n) is
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
