English

Evaluate Each of the Following Integral: ∫ B a X 1 N X 1 N + ( a + B − X ) 1 N D X , N ∈ N , N ≥ 2

Advertisements
Advertisements

Question

Evaluate each of the following integral:

\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]

Sum
Advertisements

Solution

\[\text{Let I} =\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx ........................\left( 1 \right)\]

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

\[2I = \int_a^b \frac{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx\]
\[ \Rightarrow 2I = \int_a^b dx\]
\[ \Rightarrow 2I = x_a^b = \left( b - a \right)\]
\[ \Rightarrow I = \frac{b - a}{2}\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Definite Integrals - Exercise 20.4 [Page 61]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
Exercise 20.4 | Q 10 | Page 61

RELATED QUESTIONS

\[\int\limits_4^9 \frac{1}{\sqrt{x}} dx\]

\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\]

\[\int\limits_0^{2\pi} e^{x/2} \sin\left( \frac{x}{2} + \frac{\pi}{4} \right) dx\]

\[\int_0^1 x\log\left( 1 + 2x \right)dx\]

\[\int\limits_0^1 \frac{\sqrt{\tan^{- 1} x}}{1 + x^2} dx\]

\[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\]

\[\int\limits_4^{12} x \left( x - 4 \right)^{1/3} dx\]

\[\int\limits_0^{\pi/2} \cos^5 x\ dx\]

\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]

\[\int\limits_{- a}^a \sqrt{\frac{a - x}{a + x}} dx\]

\[\int\limits_0^{\pi/2} \frac{\sin x \cos x}{\cos^2 x + 3 \cos x + 2} dx\]

\[\int_0^\frac{1}{2} \frac{1}{\left( 1 + x^2 \right)\sqrt{1 - x^2}}dx\]

\[\int_{- 2}^2 x e^\left| x \right| dx\]

If  \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]

 


Evaluate the following integral:

\[\int_{- 1}^1 \left| xcos\pi x \right|dx\]

 


\[\int\limits_0^2 \left( x^2 + 4 \right) dx\]

\[\int\limits_1^4 \left( 3 x^2 + 2x \right) dx\]

\[\int\limits_2^3 x^2 dx\]

\[\int\limits_0^2 \left( x^2 - x \right) dx\]

\[\int\limits_{- \pi/2}^{\pi/2} \cos^2 x\ dx .\]

\[\int\limits_0^{\pi/2} \sqrt{1 - \cos 2x}\ dx .\]

\[\int\limits_0^1 \frac{2x}{1 + x^2} dx\]

\[\int\limits_0^1 2^{x - \left[ x \right]} dx\]

\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals


\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\]  equals


\[\int\limits_1^e \log x\ dx =\]

\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan x} dx\]  is equal to

If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals

 


The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is 

 


Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .


\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]


\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]


\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]


Find : `∫_a^b logx/x` 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_0^1 x/((1 - x)^(3/4))  "d"x`


Choose the correct alternative:

`Γ(3/2)`


Evaluate the following:

`int ((x^2 + 2))/(x + 1) "d"x`


`int x^3/(x + 1)` is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×