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 - Mathematics

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

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
Definite Integrals
  Is there an error in this question or solution?
Chapter 20: Definite Integrals - Exercise 20.4 [Page 61]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 20 Definite Integrals
Exercise 20.4 | Q 10 | Page 61

RELATED QUESTIONS

\[\int\limits_{- 2}^3 \frac{1}{x + 7} dx\]

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

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

\[\int\limits_0^{\pi/2} \left( a^2 \cos^2 x + b^2 \sin^2 x \right) dx\]

\[\int\limits_0^1 \frac{2x + 3}{5 x^2 + 1} dx\]

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

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

\[\int_0^\frac{\pi}{4} \left( \tan x + \cot x \right)^{- 2} dx\]

\[\int\limits_0^a \sqrt{a^2 - x^2} dx\]

\[\int\limits_0^\pi \frac{1}{3 + 2 \sin x + \cos x} dx\]

\[\int_0^\frac{\pi}{4} \frac{\sin x + \cos x}{3 + \sin2x}dx\]

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

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

\[\int\limits_0^5 \frac{\sqrt[4]{x + 4}}{\sqrt[4]{x + 4} + \sqrt[4]{9 - x}} dx\]

\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\tan x}} dx\]

\[\int\limits_0^\pi x \sin x \cos^4 x\ dx\]

\[\int\limits_0^\pi x \sin^3 x\ dx\]

\[\int\limits_{- \pi/2}^{\pi/2} \sin^3 x\ dx\]

Evaluate the following integral:

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

 


\[\int\limits_0^2 e^x dx\]

\[\int\limits_a^b x\ dx\]

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

\[\int\limits_0^{\pi/4} \tan^2 x\ dx .\]

\[\int\limits_0^{\pi/2} \log \tan x\ dx .\]

If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.

 


\[\int\limits_0^\sqrt{2} \left[ x^2 \right] dx .\]

\[\int\limits_0^\pi \frac{1}{a + b \cos x} dx =\]

\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\cot}x} dx\] is

\[\int\limits_0^3 \frac{3x + 1}{x^2 + 9} dx =\]

\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^3 x} dx\]  is equal to

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

 


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


\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]


\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]


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


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


\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]


\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]


Evaluate the following:

f(x) = `{{:("c"x",", 0 < x < 1),(0",",  "otherwise"):}` Find 'c" if `int_0^1 "f"(x)  "d"x` = 2


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×