मराठी

∫ − 2 1 X + 7 D X

Advertisements
Advertisements

प्रश्न

\[\int\limits_{- 2}^3 \frac{1}{x + 7} dx\]
Advertisements

उत्तर

\[Let I = \int_{- 2}^3 \frac{1}{x + 7} d x . Then, \]
\[I = \left[ \log \left( x + 7 \right) \right]_{- 2}^3 \]
\[ \Rightarrow I = \log 10 - \log 5\]
\[ \Rightarrow I = \log \frac{10}{5} \left[ \because \log a - \log b = \log\frac{a}{b} \right]\]
\[ \Rightarrow I = \log 2\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Exercise 20.1 [पृष्ठ १६]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Exercise 20.1 | Q 2 | पृष्ठ १६

संबंधित प्रश्‍न

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

\[\int\limits_0^\infty e^{- x} dx\]

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

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

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

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

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

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

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

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


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

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

\[\int_{- \frac{\pi}{4}}^\frac{\pi}{2} \sin x\left| \sin x \right|dx\]

 


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

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

\[\int\limits_0^\pi \log\left( 1 - \cos x \right) dx\]

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

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

\[\int\limits_{- 1}^1 \left( x + 3 \right) dx\]

\[\int\limits_0^1 \left( 3 x^2 + 5x \right) dx\]

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

\[\int\limits_{- \pi/2}^{\pi/2} \log\left( \frac{a - \sin \theta}{a + \sin \theta} \right) d\theta\]

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

 


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


\[\int\limits_0^1 \frac{d}{dx}\left\{ \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \right\} dx\] is equal to

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

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

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


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


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


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


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


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


Choose the correct alternative:

`int_0^oo "e"^(-2x)  "d"x` is


Choose the correct alternative:

Γ(1) is


`int x^9/(4x^2 + 1)^6  "d"x` is equal to ______.


What is the meaning of a definite integral?


If velocity changes with time, what does a definite integral give over a time interval?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×