Advertisements
Advertisements
प्रश्न
Evaluate each of the following integral:
\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]
Advertisements
उत्तर
\[\text{Let I} =\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx ........................\left( 1 \right)\]
Then,
\[I = \int_a^b \frac{\left( a + b - x \right)^\frac{1}{n}}{\left( a + b - x \right)^\frac{1}{n} + \left[ a + b - \left( a + b - x \right) \right]^\frac{1}{n}}dx .........................\left[ \int_a^b f\left( x \right)dx = \int_a^b f\left( a + b - x \right)dx \right]\]
\[ = \int_a^b \frac{\left( a + b - x \right)^\frac{1}{n}}{\left( a + b - x \right)^\frac{1}{n} + x^\frac{1}{n}}dx ...................\left( 2 \right)\]
Adding (1) and (2), we get
\[2I = \int_a^b \frac{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx\]
\[ \Rightarrow 2I = \int_a^b dx\]
\[ \Rightarrow 2I = x_a^b = \left( b - a \right)\]
\[ \Rightarrow I = \frac{b - a}{2}\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(1/4) sqrt(1 - 4) "d"x`
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
Choose the correct alternative:
`Γ(3/2)`
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
