Advertisements
Advertisements
Question
Solution
\[Let\ I = \int_a^b \frac{f\left( x \right)}{f\left( x \right) + f\left( a + b - x \right)} d x ................(1)\]
\[ = \int_a^b \frac{f\left( a + b - x \right)}{f\left( a + b - x \right) + f\left( a + b - a - b + x \right)} d x\]
\[ = \int_a^b \frac{f\left( a + b - x \right)}{f\left( a + b - x \right) + f\left( x \right)} d x\]
\[ \therefore I = \int_a^b \frac{f\left( a + b - x \right)}{f\left( x \right) + f\left( a + b - x \right)} d x ...............(2)\]
\[\text{Adding (1) and (2) we get}\]
\[2I = \int_a^b \left[ \frac{f\left( x \right)}{f\left( x \right) + f\left( a + b - x \right)} + \frac{f\left( a + b - x \right)}{f\left( x \right) + f\left( a + b - x \right)} \right] d x\]
\[ = \int_a^b \frac{f\left( x \right) + f\left( a + b - x \right)}{f\left( x \right) + f\left( a + b - x \right)} dx\]
\[ = \left[ x \right]_a^b \]
\[ = b - a\]
\[\text{Hence, }I = \frac{b - a}{2}\]
APPEARS IN
RELATED QUESTIONS
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
Evaluate the following integral:
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
Using second fundamental theorem, evaluate the following:
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x`
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
Choose the correct alternative:
`int_0^oo "e"^(-2x) "d"x` is
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.
