हिंदी

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

Advertisements
Advertisements

प्रश्न

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

योग
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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2015-2016 (March) All India Set 1 E

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Evaluate:  `int (1+logx)/(x(2+logx)(3+logx))dx`


Evaluate: `int1/(xlogxlog(logx))dx`


Evaluate : `int_0^4(|x|+|x-2|+|x-4|)dx`


Evaluate :

`∫_0^π(4x sin x)/(1+cos^2 x) dx`


Evaluate the integral by using substitution.

`int_0^1 x/(x^2 +1)`dx


Evaluate the integral by using substitution.

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


Evaluate the integral by using substitution.

`int_1^2 (1/x- 1/(2x^2))e^(2x) dx`


Evaluate of the following integral:

(i)  \[\int x^4 dx\]

 


Evaluate of the following integral:

\[\int \log_x \text{x  dx}\] 

Evaluate: 

\[\int\sqrt{\frac{1 + \cos 2x}{2}}dx\]

Evaluate the following integral:

\[\int\limits_{- 2}^2 \left| x + 1 \right| dx\]

 


Evaluate the following integral:

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

Evaluate the following integral:

\[\int\limits_2^8 \left| x - 5 \right| dx\]

 


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 each of the following integral:

\[\int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{\tan^2 x}{1 + e^x}dx\]

 


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

Evaluate the following integral:

\[\int_{- 2}^2 \frac{3 x^3 + 2\left| x \right| + 1}{x^2 + \left| x \right| + 1}dx\]

Evaluate : \[\int\limits_{- 2}^1 \left| x^3 - x \right|dx\] .


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


Evaluate: `int_-π^π (1 - "x"^2) sin "x" cos^2 "x"  d"x"`.


Find: `int_  (3"x"+ 5)sqrt(5 + 4"x"-2"x"^2)d"x"`.


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


`int_0^(pi4) sec^4x  "d"x` = ______.


`int_0^1 x^2e^x dx` = ______.


The value of `int_0^1 (x^4(1 - x)^4)/(1 + x^2) dx` is


Evaluate: `int_0^(π/2) sin 2x tan^-1 (sin x) dx`.


Evaluate:

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×