Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_a^b \frac{f\left( x \right)}{f\left( x \right) + f\left( a + b - x \right)} d x ................(1)\]
\[ = \int_a^b \frac{f\left( a + b - x \right)}{f\left( a + b - x \right) + f\left( a + b - a - b + x \right)} d x\]
\[ = \int_a^b \frac{f\left( a + b - x \right)}{f\left( a + b - x \right) + f\left( x \right)} d x\]
\[ \therefore I = \int_a^b \frac{f\left( a + b - x \right)}{f\left( x \right) + f\left( a + b - x \right)} d x ...............(2)\]
\[\text{Adding (1) and (2) we get}\]
\[2I = \int_a^b \left[ \frac{f\left( x \right)}{f\left( x \right) + f\left( a + b - x \right)} + \frac{f\left( a + b - x \right)}{f\left( x \right) + f\left( a + b - x \right)} \right] d x\]
\[ = \int_a^b \frac{f\left( x \right) + f\left( a + b - x \right)}{f\left( x \right) + f\left( a + b - x \right)} dx\]
\[ = \left[ x \right]_a^b \]
\[ = b - a\]
\[\text{Hence, }I = \frac{b - a}{2}\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^4 x dx\]
Evaluate the following integrals as the limit of the sum:
`int_0^1 x^2 "d"x`
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
`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:
Find: `int logx/(1 + log x)^2 dx`
