मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Evaluate: ∫38(1-x)2x2+(11-x)2 dx - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

उत्तर

Let I = `int_3^8 (11 - x)^2/(x^2 + (11 - x)^2)  "d"x`   .......(i)

= `int_3^8 ([11 - (1 - x)]^2)/((11 - x)^2 + [11 - (11 - x)]2)  "d"x`      ........`[∵ int_"a"^"b" "f"(x)"d"x = int_"a"^"b" "f"("a" + "b" - x)"d"x]`

∴ I = `int_3^8 x^2/((11 - x)^2 + x^2)  "d"x` .......(ii)

Adding (i) and (ii), we get

2I = `int_3^8 (11 - x)^2/(x^2 + (11 - x)^2)  "d"x + int_3^8 x^2/((11 - x)^2 + x^2)  "d"x`

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

∴ 2I = `int_3^8 1. "d"x`

∴ I= `1/2[x]_3^8`

∴ I = `1/2(8 -3)`

∴ I =`5/2`

shaalaa.com
Methods of Evaluation and Properties of Definite Integral
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 2.4: Definite Integration - Short Answers II

संबंधित प्रश्‍न

Evaluate: `int_0^(π/4) cot^2x.dx`


Evaluate: `int_0^π sin^3x (1 + 2cosx)(1 + cosx)^2.dx`


Choose the correct option from the given alternatives : 

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


If `int_0^1 ("d"x)/(sqrt(1 + x) - sqrt(x)) = "k"/3`, then k is equal to ______.


`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_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_0^1 1/sqrt(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_0^(pi/4) (tan^3x)/(1 + cos 2x)  "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^(pi/2) 1/(5 + 4cos x)  "d"x`


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


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


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


Evaluate: `int_0^(π/4) sec^4 x  dx`


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


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/2)^(π/2) |sinx|dx`


Evaluate `int_(π/6)^(π/3) cos^2x  dx`


Evaluate:

`int_-4^5 |x + 3|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` = ______.


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)`


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


If `int_0^π f(sinx)dx = kint_0^π f(sinx)dx`, then find the value of k.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×