मराठी

1 ∫ 0 1 √ 1 + X − √ X D X

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[Let I = \int_0^1 \frac{1}{\sqrt{1 + x} - \sqrt{x}} d x . Then, \]
\[I = \int_0^1 \left( \frac{1}{\sqrt{1 + x} - \sqrt{x}} \times \frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} + \sqrt{x}} \right) d x\]
\[ \Rightarrow I = \int_0^1 \frac{\sqrt{1 + x} + \sqrt{x}}{1 + x - x} d x\]
\[ \Rightarrow I = \int_0^1 \left( \sqrt{1 + x} + \sqrt{x} \right) dx\]
\[ \Rightarrow I = \left[ \frac{2}{3} \left( 1 + x \right)^\frac{3}{2} + \frac{2}{3} x^\frac{3}{2} \right]_0^1 \]
\[ \Rightarrow I = \frac{2}{3} \times 2\sqrt{2} + \frac{2}{3} - \frac{2}{3}\]
\[ \Rightarrow I = \frac{4\sqrt{2}}{3}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Exercise 20.1 [पृष्ठ १७]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Exercise 20.1 | Q 54 | पृष्ठ १७

संबंधित प्रश्‍न

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

\[\int\limits_{\pi/3}^{\pi/4} \left( \tan x + \cot x \right)^2 dx\]

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

\[\int\limits_0^4 \frac{1}{\sqrt{4x - x^2}} dx\]

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

\[\int_0^1 \frac{1}{1 + 2x + 2 x^2 + 2 x^3 + x^4}dx\]

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

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

\[\int_0^\frac{\pi}{2} \frac{\cos^2 x}{1 + 3 \sin^2 x}dx\]

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

 


\[\int\limits_0^\pi x \cos^2 x\ dx\]

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

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

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

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

 


If f(2a − x) = −f(x), prove that

\[\int\limits_0^{2a} f\left( x \right) dx = 0 .\]

If f (x) is a continuous function defined on [0, 2a]. Then, prove that

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

 


\[\int\limits_3^5 \left( 2 - x \right) dx\]

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

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

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

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

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{4} \tan\ xdx\]

 


Evaluate each of the following integral:

\[\int_e^{e^2} \frac{1}{x\log x}dx\]

If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.

 

 


Evaluate : 

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

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


`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`


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

\[\lim_{n \to \infty} \left\{ \frac{1}{2n + 1} + \frac{1}{2n + 2} + . . . + \frac{1}{2n + n} \right\}\] is equal to

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


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


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


\[\int\limits_{- 1}^1 e^{2x} dx\]


Evaluate the following using properties of definite integral:

`int_(- pi/2)^(pi/2) sin^2theta  "d"theta`


Evaluate the following integrals as the limit of the sum:

`int_0^1 x^2  "d"x`


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


Find `int sqrt(10 - 4x + 4x^2)  "d"x`


The value of `int_2^3 x/(x^2 + 1)`dx is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×