हिंदी

Evaluate: ∫01|x| dx - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Evaluate: `int_0^1 |x|  "d"x`

योग
Advertisements

उत्तर

`int_0^1 |x|  "d"x - int_0^1 x  "d"x`   ......[|x| = x, x ≥ 0]

= `[x^2/2]_0^(-1)`

= `1/2(1^2 - 0)`

= `1/2`

shaalaa.com
Methods of Evaluation and Properties of Definite Integral
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2.4: Definite Integration - Very Short Answers

संबंधित प्रश्न

Evaluate: `int_0^1 (x^2 - 2)/(x^2 + 1).dx`


Evaluate: `int_0^oo xe^-x.dx`


Choose the correct option from the given alternatives : 

`int_0^(pi/2) (sin^2x*dx)/(1 + cosx)^2` = ______.


`int_0^(x/4) sqrt(1 + sin 2x)  "d"x` =


`int_(pi/5)^((3pi)/10)  sinx/(sinx + cosx)  "d"x` =


Let I1 = `int_"e"^("e"^2)  1/logx  "d"x` and I2 = `int_1^2 ("e"^x)/x  "d"x` then 


Evaluate: `int_0^(pi/4) sec^2 x  "d"x`


Evaluate: `int_0^1 1/sqrt(1 - x^2)  "d"x`


Evaluate: `int_0^1(x + 1)^2  "d"x`


Evaluate: `int_0^(pi/2) sqrt(1 - cos 4x)  "d"x`


Evaluate:

`int_0^(pi/2) cos^3x  dx`


Evaluate: `int_0^pi cos^2 x  "d"x`


Evaluate: `int_0^(pi/2) (sin^2x)/(1 + cos x)^2 "d"x`


Evaluate: `int_3^8 (11 - x)^2/(x^2 + (11 - x)^2)  "d"x`


Evaluate: `int_(-4)^2 1/(x^2 + 4x + 13)  "d"x`


Evaluate: `int_0^1 x* tan^-1x  "d"x`


Evaluate: `int_0^(1/sqrt(2)) (sin^-1x)/(1 - x^2)^(3/2)  "d"x`


Evaluate: `int_0^(pi/4) sec^4x  "d"x`


Evaluate: `int_0^(pi/2) 1/(5 + 4cos x)  "d"x`


Evaluate: `int_0^(pi/2) cos x/((1 + sinx)(2 + sinx))  "d"x`


Evaluate: `int_0^"a" 1/(x + sqrt("a"^2 - x^2))  "d"x`


Evaluate: `int_0^3 x^2 (3 - x)^(5/2)  "d"x`


Evaluate: `int_0^1 "t"^2 sqrt(1 - "t")  "dt"`


Evaluate: `int_0^(pi/4)  (sec^2x)/(3tan^2x + 4tan x + 1)  "d"x`


Evaluate: `int_(1/sqrt(2))^1  (("e"^(cos^-1x))(sin^-1x))/sqrt(1 - x^2)  "d"x`


Evaluate: `int_0^pi x*sinx*cos^2x* "d"x`


Evaluate: `int_(-1)^1 (1 + x^2)/(9 - x^2)  "d"x`


Evaluate: `int_0^1 (1/(1 + x^2)) sin^-1 ((2x)/(1 + x^2))  "d"x`


Evaluate: `int_0^(pi/4) log(1 + tanx)  "d"x`


Evaluate: `int_0^pi 1/(3 + 2sinx + cosx)  "d"x`


`int_0^(π/2) sin^6x cos^2x.dx` = ______.


If `int_2^e [1/logx - 1/(logx)^2].dx = a + b/log2`, then ______.


Evaluate:

`int_(π/4)^(π/2) cot^2x  dx`.


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


Evaluate:

`int_0^(π/2) sin^8x  dx`


Evaluate:

`int_(π/6)^(π/3) (root(3)(sinx))/(root(3)(sinx) + root(3)(cosx))dx`


Evaluate:

`int_0^(π/2) (sin 2x)/(1 + sin^4x)dx`


Evaluate:

`int_0^(π/2) sinx/(1 + cosx)^3 dx`


Prove that: `int_0^1 logx/sqrt(1 - x^2)dx = π/2 log(1/2)`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×