मराठी

1 ∫ − 1 X | X | D X .

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[\left| x \right| = \begin{cases} - x &,& - 1 < x < 0\\ x &,& 0 < x < 1\end{cases}\]
\[ \therefore x\left| x \right| = \begin{cases} - x^2 &,& - 1 < x < 0\\ x^2 &,& 0 < x < 1\end{cases}\]
\[Now\, \int_{- 1}^1 x\left| x \right| d x\]
\[ = \int_{- 1}^0 - x^2 dx + \int_0^1 x^2 dx\]
\[ = - \int_{- 1}^0 x^2 dx + \int_0^1 x^2 dx\]
\[ = - \left[ \frac{x^3}{3} \right]_{- 1}^0 + \left[ \frac{x^3}{3} \right]_0^1 \]
\[ = - \left( 0 + \frac{1}{3} \right) + \left( \frac{1}{3} - 0 \right)\]
\[ = 0 - \frac{1}{3} + \frac{1}{3} - 0\]
\[ = 0\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Very Short Answers [पृष्ठ ११५]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Very Short Answers | Q 19 | पृष्ठ ११५

संबंधित प्रश्‍न

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

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

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

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

\[\int_0^1 \frac{1}{1 + 2x + 2 x^2 + 2 x^3 + x^4}dx\]

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

\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]


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

\[\int\limits_0^\pi x \sin^3 x\ dx\]

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

 


If `f` is an integrable function such that f(2a − x) = f(x), then prove that

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

 


If f(x) is a continuous function defined on [−aa], then prove that 

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

\[\int\limits_0^3 \left( x + 4 \right) dx\]

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

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

\[\int\limits_0^2 \sqrt{4 - x^2} dx\]

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

\[\int\limits_0^1 e^\left\{ x \right\} dx .\]

\[\int\limits_0^3 \frac{3x + 1}{x^2 + 9} dx =\]

\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{\sin 2x} dx\]  is equal to

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


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

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


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


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


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


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


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


Using second fundamental theorem, evaluate the following:

`int_0^(1/4) sqrt(1 - 4)  "d"x`


Evaluate the following:

f(x) = `{{:("c"x",", 0 < x < 1),(0",",  "otherwise"):}` Find 'c" if `int_0^1 "f"(x)  "d"x` = 2


Evaluate the following:

`Γ (9/2)`


Choose the correct alternative:

If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x)  "d"x + int_"c"^"b" f(x)  "d"x` is


Choose the correct alternative:

Using the factorial representation of the gamma function, which of the following is the solution for the gamma function Γ(n) when n = 8 is


Choose the correct alternative:

`int_0^oo x^4"e"^-x  "d"x` is


`int x^3/(x + 1)` is equal to ______.


What is the result of a definite integral?


Evaluate \[\int_{0}^{1}x\,dx\].


What is an antiderivative of \[2x+1\]?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×