हिंदी

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

प्रश्न

Evaluate each of the following integral:

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

योग
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

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

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Definite Integrals - Exercise 20.4 [पृष्ठ ६१]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Exercise 20.4 | Q 10 | पृष्ठ ६१

संबंधित प्रश्न

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

\[\int\limits_{- \pi/4}^{\pi/4} \frac{1}{1 + \sin x} dx\]

\[\int\limits_0^{\pi/2} \sqrt{1 + \sin x}\ dx\]

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

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

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

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

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

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

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

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

\[\int_0^\frac{\pi}{2} \frac{\tan x}{1 + m^2 \tan^2 x}dx\]

Evaluate the following integral:

\[\int\limits_{- 3}^3 \left| x + 1 \right| dx\]

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

\[\int_0^2 2x\left[ x \right]dx\]

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

 


If `f` is an integrable function such that f(2a − x) = f(x), then prove that

\[\int\limits_0^{2a} f\left( x \right) dx = 2 \int\limits_0^a f\left( x \right) dx\]

 


\[\int\limits_0^2 \left( 3 x^2 - 2 \right) dx\]

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

\[\int\limits_2^3 \frac{1}{x}dx\]

\[\int\limits_0^{15} \left[ x \right] dx .\]

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

\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals


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

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


\[\int\limits_0^1 \cos^{- 1} x dx\]


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


\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]


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


\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]


\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]


Using second fundamental theorem, evaluate the following:

`int_0^1 x"e"^(x^2)  "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_(- pi/4)^(pi/4) x^3 cos^3 x  "d"x`


Evaluate the following using properties of definite integral:

`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x)  "d"x`


Choose the correct alternative:

`int_0^oo "e"^(-2x)  "d"x` is


If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.


`int (x + 3)/(x + 4)^2 "e"^x  "d"x` = ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×