मराठी

2 ∫ 1 E 2 X ( 1 X − 1 2 X 2 ) D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[Let\ I \int_1^2 e^{2x} \left( \frac{1}{x} - \frac{1}{2 x^2} \right) d\ x\ . Then, \]
\[I = \int_1^2 e^{2x} \frac{1}{x} - \int_1^2 e^{2x} \frac{1}{2 x^2} dx\]
\[\text{Integrating first term by parts}\]
\[ \Rightarrow I = \left\{ \left[ \frac{e^{2x}}{2x} \right]_1^2 - \int_1^2 - e^{2x} \frac{1}{2 x^2} \right\} - \int_1^2 e^{2x} \frac{1}{2 x^2} dx\]
\[ \Rightarrow I = \left[ \frac{e^{2x}}{2x} \right]_1^2 \]
\[ \Rightarrow I = \frac{e^4}{4} - \frac{e^2}{2}\]
\[ \Rightarrow I = \frac{e^4 - 2 e^2}{4}\]

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - Exercise 20.1 [पृष्ठ १७]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
Exercise 20.1 | Q 58 | पृष्ठ १७

संबंधित प्रश्‍न

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

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

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

\[\int\limits_0^{\pi/2} 2 \sin x \cos x \tan^{- 1} \left( \sin x \right) dx\]

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

Evaluate the following integral:

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

\[\int_0^{2\pi} \cos^{- 1} \left( \cos x \right)dx\]

Evaluate each of the following integral:

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

 


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

 


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

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

 


Evaluate the following integral:

\[\int_{- 1}^1 \left| xcos\pi x \right|dx\]

 


Prove that:

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

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

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

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

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

Evaluate each of the following  integral:

\[\int_0^1 x e^{x^2} dx\]

 


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

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

`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`


The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]

 


The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is


Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .


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


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


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


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


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


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


\[\int\limits_0^\pi \cos 2x \log \sin x dx\]


\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]


Evaluate the following integrals as the limit of the sum:

`int_0^1 (x + 4)  "d"x`


Choose the correct alternative:

`int_0^1 (2x + 1)  "d"x` is


Choose the correct alternative:

The value of `int_(- pi/2)^(pi/2) cos  x  "d"x` is


If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.


The value of `int_2^3 x/(x^2 + 1)`dx is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×