मराठी

A ∫ 0 √ X √ X + √ a − X D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

उत्तर

\[Let, I = \int_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} d x ...............(1)\]

\[ = \int_0^a \frac{\sqrt{a - x}}{\sqrt{a - x} + \sqrt{a - a + x}} d x\]

\[ = \int_0^a \frac{\sqrt{a - x}}{\sqrt{a - x} + \sqrt{x}} d x\]

\[ \Rightarrow I = \int_0^a \frac{\sqrt{a - x}}{\sqrt{x} + \sqrt{a - x}} d x.................(2)\]

Adding (1) and (2)

\[2I = \int_0^a \left[ \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} + \frac{\sqrt{a - x}}{\sqrt{x} + \sqrt{a - x}} \right] d x\]

\[ = \int_0^a dx\]

\[ = \left[ x \right]_0^a \]

\[ = a\]

\[\text{Hence, }I = \frac{a}{2}\]

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - Revision Exercise [पृष्ठ १२२]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
Revision Exercise | Q 39 | पृष्ठ १२२

संबंधित प्रश्‍न

\[\int\limits_{\pi/6}^{\pi/4} cosec\ x\ dx\]

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

\[\int\limits_1^e \frac{\log x}{x} dx\]

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

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

\[\int\limits_1^2 e^{2x} \left( \frac{1}{x} - \frac{1}{2 x^2} \right) dx\]

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

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

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


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

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

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

\[\int\limits_0^9 f\left( x \right) dx, where f\left( x \right) \begin{cases}\sin x & , & 0 \leq x \leq \pi/2 \\ 1 & , & \pi/2 \leq x \leq 3 \\ e^{x - 3} & , & 3 \leq x \leq 9\end{cases}\]

\[\int_{- \frac{\pi}{2}}^\pi \sin^{- 1} \left( \sin x \right)dx\]

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

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

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

\[\int\limits_0^5 \left( x + 1 \right) dx\]

\[\int\limits_0^4 \frac{1}{\sqrt{16 - x^2}} dx .\]

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

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

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{4} \sin2xdx\]

If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.


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

The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is 


The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .


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


Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .


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


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


\[\int\limits_0^{15} \left[ x^2 \right] dx\]


\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]


Using second fundamental theorem, evaluate the following:

`int_0^1 "e"^(2x)  "d"x`


If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`


Choose the correct alternative:

`int_(-1)^1 x^3 "e"^(x^4)  "d"x` is


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 x^2/(x^4 + 3x^2 + 2) "d"x`


Verify the following:

`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`


Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×