मराठी

3 ∫ 0 3 X + 1 X 2 + 9 D X = - Mathematics

Advertisements
Advertisements

प्रश्न

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

पर्याय

  • \[\frac{\pi}{12} + \log\left( 2\sqrt{2} \right)\]
  • \[\frac{\pi}{2} + \log\left( 2\sqrt{2} \right)\]
  • \[\frac{\pi}{6} + \log\left( 2\sqrt{2} \right)\]
  • \[\frac{\pi}{3} + \log\left( 2\sqrt{2} \right)\]

MCQ
Advertisements

उत्तर

\[\frac{\pi}{12} + \log\left( 2\sqrt{2} \right)\]

\[\text{We have}, \]
\[I = \int_0^3 \frac{3x + 1}{x^2 + 9} d x\]
\[I = \int_0^3 \frac{3x}{x^2 + 9}dx + \int_0^3 \frac{1}{x^2 + 9}dx\]
\[ I_1 = \int_0^3 \frac{3x}{x^2 + 9}dx and I_2 = \int_0^3 \frac{1}{x^2 + 9}dx\]
\[\text{Putting} x^2 + 9 = t in I_1 \]
\[ \Rightarrow 2x\ dx = dt\]
\[ \Rightarrow x\ dx = \frac{dt}{2}\]
\[When\ x \to 0; t \to 9\]
\[and\ x \to 3; t \to 18\]
\[ \therefore I = \int_9^{18} \frac{3 dt}{2 t} + \int_0^3 \frac{1}{x^2 + 9}dx\]
\[ = \frac{3}{2} \int_9^{18} \frac{dt}{t} + \int_0^3 \frac{1}{x^2 + 3^2}dx\]
\[ = \frac{3}{2} \left[ \log\left( t \right) \right]_9^{18} + \frac{1}{3} \left[ \tan^{- 1} \left( \frac{x}{3} \right) \right]_0^3 \]
\[ = \frac{3}{2}\left[ \log18 - \log9 \right] + \frac{1}{3}\left( \frac{\pi}{4} - 0 \right)\]
\[ = \frac{3}{2}\left[ \log\frac{18}{9} \right] + \frac{\pi}{12}\]
\[ = \frac{3}{2}\left[ \log 2 \right] + \frac{\pi}{12}\]
\[ = \log\left( \sqrt{8} \right) + \frac{\pi}{12}\]
\[ = \log\left( 2\sqrt{2} \right) + \frac{\pi}{12}\]
\[ = \frac{\pi}{12} + \log\left( 2\sqrt{2} \right)\]

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - MCQ [पृष्ठ ११८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
MCQ | Q 16 | पृष्ठ ११८

संबंधित प्रश्‍न

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

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

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

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

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

\[\int_0^1 x\log\left( 1 + 2x \right)dx\]

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

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

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

\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]


\[\int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx\]

 


Evaluate each of the following integral:

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


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

 


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

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

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

\[\int\limits_1^4 \left( x^2 - x \right) dx\]

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

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

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

\[\int\limits_0^1 e^\left\{ x \right\} dx .\]

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

\[\int\limits_{- 1}^1 \left| 1 - x \right| dx\]  is equal to

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


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


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


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


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


\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]


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


\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]


\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]


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


Using second fundamental theorem, evaluate the following:

`int_0^1 "e"^(2x)  "d"x`


Using second fundamental theorem, evaluate the following:

`int_0^1 x"e"^(x^2)  "d"x`


Evaluate the following:

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


Evaluate the following using properties of definite integral:

`int_(-1)^1 log ((2 - x)/(2 + x))  "d"x`


Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`


Verify the following:

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×