Advertisements
Advertisements
Question
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
Advertisements
Solution
\[\int_a^b xf\left( x \right)dx\]
\[ = \int_a^b \left( a + b - x \right)f\left( a + b - x \right)dx ..................\left[ \int_a^b f\left( x \right)dx = \int_a^b f\left( a + b - x \right)dx \right]\]
\[ = \int_a^b \left( a + b - x \right)f\left( x \right)dx ..................\left[ f\left( a + b - x \right) = f\left( x \right) \right]\]
\[ \therefore \int_a^b xf\left( x \right)dx = \int_a^b \left( a + b \right)f\left( x \right)dx - \int_a^b xf\left( x \right)dx\]
\[\Rightarrow \int_a^b xf\left( x \right)dx + \int_a^b xf\left( x \right)dx = \left( a + b \right) \int_a^b f\left( x \right)dx\]
\[ \Rightarrow 2 \int_a^b xf\left( x \right)dx = \left( a + b \right) \int_a^b f\left( x \right)dx\]
\[ \Rightarrow \int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
APPEARS IN
RELATED QUESTIONS
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
Solve each of the following integral:
If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.
Evaluate :
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
Evaluate the following:
`Γ (9/2)`
Choose the correct alternative:
If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x) "d"x + int_"c"^"b" f(x) "d"x` is
