मराठी

3 ∫ 2 √ X √ 5 − X + √ X D X

Advertisements
Advertisements

प्रश्न

\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]

बेरीज
Advertisements

उत्तर

\[Let, I = \int_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} d x ...................(1)\]

\[ = \int_2^3 \frac{\sqrt{5 - x}}{\sqrt{5 - 5 + x} + \sqrt{5 - x}} d x \]

\[ = \int_2^3 \frac{\sqrt{5 - x}}{\sqrt{x} + \sqrt{5 - x}} d x ...................(2)\]

Adding (1) and (2)

\[ 2I = \int_2^3 \left[ \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} + \frac{\sqrt{5 - x}}{\sqrt{x} + \sqrt{5 - x}} \right] d x\]

\[ = \int_2^3 \frac{\sqrt{5 - x} + \sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]

\[ = \int_2^3 dx \]

\[ = \left[ x \right]_2^3 \]

\[ = 3 - 1 = 1\]

\[Hence, I = \frac{1}{2}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Revision Exercise [पृष्ठ १२२]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Revision Exercise | Q 52 | पृष्ठ १२२

संबंधित प्रश्‍न

\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]


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

\[\int\limits_2^4 \frac{x}{x^2 + 1} dx\]

\[\int\limits_1^2 \frac{3x}{9 x^2 - 1} dx\]

\[\int\limits_0^{\pi/4} \left( \sqrt{\tan}x + \sqrt{\cot}x \right) dx\]

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

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

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

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

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

\[\int_\frac{1}{3}^1 \frac{\left( x - x^3 \right)^\frac{1}{3}}{x^4}dx\]

\[\int_0^\frac{\pi}{4} \frac{\sin^2 x \cos^2 x}{\left( \sin^3 x + \cos^3 x \right)^2}dx\]

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

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

Evaluate each of the following integral:

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


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

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

Prove that:

\[\int_0^\pi xf\left( \sin x \right)dx = \frac{\pi}{2} \int_0^\pi f\left( \sin x \right)dx\]

Solve each of the following integral:

\[\int_2^4 \frac{x}{x^2 + 1}dx\]

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

 

 


Evaluate : 

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

\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.  

 

\[\int\limits_{- \pi/2}^{\pi/2} \sin\left| x \right| dx\]  is equal to

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


The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is 

 


If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\]  then the value of I10 + 90I8 is

 


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

 


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


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


Evaluate the following integrals :-

\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]


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


\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]


\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]


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


Evaluate the following:

`int_(-1)^1 "f"(x)  "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x  < 0):}`


Evaluate the following:

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


Choose the correct alternative:

`Γ(3/2)`


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×