मराठी

Evaluate: ∫ 5 1 { | X − 1 | + | X − 2 | + | X − 3 | } D X .

Advertisements
Advertisements

प्रश्न

Evaluate: `int_1^5{|"x"-1|+|"x"-2|+|"x"-3|}d"x"`.

बेरीज
Advertisements

उत्तर

= `int_1^5 {| x - 1| + | x - 2| + |x - 3|} dx`

= `int_1^5 (x - 1)dx + int_1^2 (2 - x)dx + int_2^5 (x - 2) dx + int_1^3 (3 - x) dx + int_3^5 ( x - 3) dx`

= `[x^2/2 - x]_1^5 + [2x - x^2/2]_1^2 + [x^2/2 - 2x]_2^5 + [3x - x^2/2]_1^3 + [x^2/3 - 3x]_3^5`

= 17

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2015-2016 (March) All India Set 1 E

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Evaluate :

`∫_(-pi)^pi (cos ax−sin bx)^2 dx`


If `int_0^a1/(4+x^2)dx=pi/8` , find the value of a.


Evaluate the integral by using substitution.

`int_0^2 xsqrt(x+2)`  (Put x + 2 = `t^2`)


The value of the integral `int_(1/3)^4 ((x- x^3)^(1/3))/x^4` dx is ______.


If `f(x) = int_0^pi t sin  t  dt`, then f' (x) is ______.


Evaluate of the following integral: 

\[\int x^\frac{5}{4} dx\]

Evaluate of the following integral:

\[\int\frac{1}{\sqrt[3]{x^2}}dx\]

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

Evaluate the following integral:

\[\int\limits_{- 4}^4 \left| x + 2 \right| dx\]

Evaluate the following integral:

\[\int\limits_0^3 \left| 3x - 1 \right| dx\]

 


Evaluate the following integral:

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

Evaluate the following integral:

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

Evaluate the following integral:

\[\int\limits_0^4 \left| x - 1 \right| dx\]

Evaluate each of the following integral:

\[\int_0^{2\pi} \frac{e^\ sin x}{e^\ sin x + e^{- \ sin x}}dx\]

 


Evaluate each of the following integral:

\[\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sqrt{\tan x}}{\sqrt{\tan x} + \sqrt{\cot x}}dx\]

Evaluate 

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


Find : \[\int\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}dx\] .


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


`int_(pi/5)^((3pi)/10) [(tan x)/(tan x + cot x)]`dx = ?


`int_0^1 x(1 - x)^5 "dx" =` ______.


`int_0^3 1/sqrt(3x - x^2)"d"x` = ______.


`int_0^(pi4) sec^4x  "d"x` = ______.


Find: `int (dx)/sqrt(3 - 2x - x^2)`


Evaluate:

`int (1 + cosx)/(sin^2x)dx`


Under Method: Changing the Limits, what does the lower limit \[a\] become?


What is the value of \[\int_{0}^{1}\frac{\tan^{-1}x}{1+x^2}\,dx\]?


When an integral is continued in the new variable, what must be done to its limits?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×