मराठी

2 ∫ 1 1 X 2 E − 1 / X D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

उत्तर

\[\int_1^2 \frac{1}{x^2} e^\frac{- 1}{x} d x\]

\[Let \frac{- 1}{x} = t, then \frac{1}{x^2} dx = dt\]

\[\text{When, }x \to 1 ; t \to - 1\]

\[\text{And }x \to 2 ; t \to \frac{- 1}{2}\]

Therefore the integral becomes

\[ \int_{- 1}^\frac{- 1}{2} e^t d t\]

\[ = \left[ e^t \right]_{- 1}^\frac{- 1}{2} \]

\[ = e^\frac{- 1}{2} - e^{- 1} \]

\[ = \frac{\sqrt{e} - 1}{e}\]

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - Revision Exercise [पृष्ठ १२१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
Revision Exercise | Q 18 | पृष्ठ १२१

संबंधित प्रश्‍न

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

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

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

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

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

 


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

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

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

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

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

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

\[\int\limits_0^{\pi/2} \frac{1}{5 \cos x + 3 \sin x} dx\]

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

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

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

\[\int\limits_0^{\pi/2} \frac{\sin \theta}{\sqrt{1 + \cos \theta}} d\theta\]

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

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

\[\int\limits_0^\pi 5 \left( 5 - 4 \cos \theta \right)^{1/4} \sin \theta\ d \theta\]

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

\[\int_0^\pi \cos x\left| \cos x \right|dx\]

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

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

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

If f (x) is a continuous function defined on [0, 2a]. Then, prove that

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

 


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

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

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{4} \tan\ xdx\]

 


Evaluate each of the following integral:

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

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

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

 


\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^3 x} dx\]  is equal to

\[\int\limits_0^1 \left| 2x - 1 \right| dx\]


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


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


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


Using second fundamental theorem, evaluate the following:

`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`


Evaluate the following:

`int ((x^2 + 2))/(x + 1) "d"x`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×