मराठी

∫ 2 − 2 X E | X | D X

Advertisements
Advertisements

प्रश्न

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

उत्तर

Consider

\[f\left( x \right) = x e^\left| x \right|\]

Now,

\[f\left( - x \right) = \left( - x \right) e^\left| - x \right| = - x e^\left| x \right| = - f\left( x \right)\]

⇒ f(x) is an odd function.

\[\therefore \int_{- 2}^2 x e^\left| x \right| dx = 0\]

\[\left[ \int_{- a}^a f\left( x \right)dx = \begin{cases}2 \int_0^a f\left( x \right)dx, & \text{if }f\left( - x \right) = f\left( x \right) \\ 0, & \text{if }f\left( - x \right) = - f\left( x \right)\end{cases} \right]\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Exercise 20.3 [पृष्ठ ५६]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Exercise 20.3 | Q 21 | पृष्ठ ५६

संबंधित प्रश्‍न

\[\int\limits_4^9 \frac{1}{\sqrt{x}} dx\]

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

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

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

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

\[\int_0^\frac{\pi}{4} \left( \tan x + \cot x \right)^{- 2} dx\]

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

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

\[\int\limits_0^{\pi/2} \frac{dx}{a \cos x + b \sin x}a, b > 0\]

Evaluate each of the following integral:

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

 


\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx, 0 < \alpha < \pi\]

Evaluate the following integral:

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

 


If f is an integrable function, show that

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

 


Prove that:

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

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

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

\[\int\limits_2^3 x^2 dx\]

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

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

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

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


The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\]  is 


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

 


\[\lim_{n \to \infty} \left\{ \frac{1}{2n + 1} + \frac{1}{2n + 2} + . . . + \frac{1}{2n + n} \right\}\] is equal to

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


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


The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is 

 


The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is 


Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .


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


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


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


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


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


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


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


If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`


Choose the correct alternative:

`int_0^oo x^4"e"^-x  "d"x` is


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×