हिंदी

∫01x2-2x2+1 dx =

Advertisements
Advertisements

प्रश्न

`int_0^1 (x^2 - 2)/(x^2 + 1)  "d"x` =

विकल्प

  • `1 - (3pi)/4`

  • `2 - (3pi)/4`

  • `1 + (3pi)/4`

  • `2 + (3pi)/4`

MCQ
Advertisements

उत्तर

`1 - (3pi)/4`

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

संबंधित प्रश्न

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


Evaluate: `int_0^(pi/2) x sin x.dx`


Evaluate the following:

`int_0^a (1)/(x + sqrt(a^2 - x^2)).dx`


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


`int_0^4 1/sqrt(4x - x^2)  "d"x` =


`int_0^(pi/2) log(tanx)  "d"x` =


Evaluate: `int_(- pi/4)^(pi/4) x^3 sin^4x  "d"x`


Evaluate: `int_0^1 1/(1 + x^2)  "d"x`


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


Evaluate: `int_0^1 |x|  "d"x`


Evaluate: `int_1^2 x/(1 + x^2)  "d"x`


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


Evaluate: `int_0^(pi/2)  (sin2x)/(1 +  sin^2x)  "d"x`


Evaluate: `int_(pi/6)^(pi/3) sin^2 x  "d"x`


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


Evaluate: `int_1^3 (cos(logx))/x  "d"x`


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


Evaluate: `int_0^9 sqrt(x)/(sqrt(x) + sqrt(9 - x)  "d"x`


Evaluate: `int_0^1 1/sqrt(3 + 2x - x^2)  "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/2) 1/(5 + 4cos x)  "d"x`


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


Evaluate: `int_0^1 (log(x + 1))/(x^2 + 1)  "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^(pi/4)  (cos2x)/(1 + cos 2x + sin 2x)  "d"x`


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


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


Evaluate:

`int_(-π/2)^(π/2) |sinx|dx`


Evaluate:

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


`int_0^1 x^2/(1 + x^2)dx` = ______.


Find the value of ‘a’ if `int_2^a (x + 1)dx = 7/2`


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


Evaluate `int_(-π/2)^(π/2) sinx/(1 + cos^2x)dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×