मराठी

2 ∫ 1 Log E [ X ] D X . - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\limits_1^2 \log_e \left[ x \right] dx .\]
बेरीज
Advertisements

उत्तर

\[\text{We have}, \]
\[I = \int\limits_1^2 \log_e \left[ x \right] dx\]
\[\text{We know that}, \]
\[\left[ x \right] = 1\text{, when }1 < x < 2\]
\[ \therefore I = \int\limits_1^2 \log_e 1 dx\]
\[I = \int\limits_1^2 \left( 0 \right) dx\]
\[ = 0\]

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - Very Short Answers [पृष्ठ ११६]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
Very Short Answers | Q 43 | पृष्ठ ११६

संबंधित प्रश्‍न

\[\int\limits_0^{\pi/4} \sec x dx\]

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

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

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

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

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

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

\[\int\limits_0^{\pi/4} \frac{\tan^3 x}{1 + \cos 2x} dx\]

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

\[\int\limits_0^1 \frac{24 x^3}{\left( 1 + x^2 \right)^4} dx\]

\[\int\limits_0^{\pi/6} \cos^{- 3} 2 \theta \sin 2\ \theta\ d\ \theta\]

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

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

Evaluate each of the following integral:

\[\int_0^{2\pi} \log\left( \sec x + \tan x \right)dx\]

 


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

If f(2a − x) = −f(x), prove that

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

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

\[\int\limits_a^b \cos\ x\ dx\]

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

\[\int\limits_0^3 \left( 2 x^2 + 3x + 5 \right) dx\]

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

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

Solve each of the following integral:

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

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

\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan x} dx\]  is equal to

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


\[\int\limits_0^{\pi/2} \sin\ 2x\ \log\ \tan x\ dx\]  is equal to 

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


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


\[\int\limits_0^{\pi/4} \tan^4 x dx\]


\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]


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


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


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


Evaluate the following:

`Γ (9/2)`


Choose the correct alternative:

Γ(1) is


Choose the correct alternative:

`int_0^oo x^4"e"^-x  "d"x` is


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


Find: `int logx/(1 + log x)^2 dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×