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

Evaluate: eed∫01exex-1 dx - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

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

बेरीज

उत्तर

`int_0^1 "e"^x/sqrt("e"^x - 1)  "d"x = [2sqrt("e"^x - 1)]_0^1`        ......`[∵ int ("f'"(x))/sqrt("f"(x))  "d"x = 2sqrt("f"(x)) + "c"]`

= `2(sqrt("e"^1  1) - sqrt("e"^0 - 1))`

= `2(sqrt("e" - 1) - sqrt(1 - 1))`

= `2sqrt("e" - 1)`

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

संबंधित प्रश्‍न

Evaluate: `int_0^(π/4) cot^2x.dx`


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


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


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_0^1 (x^2 - 2)/(x^2 + 1)  "d"x` =


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


`int_0^4 1/sqrt(4x - x^2)  "d"x` =


`int_0^(pi/2) log(tanx)  "d"x` =


Evaluate: `int_(pi/6)^(pi/3) cosx  "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_0^1 |x|  "d"x`


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


Evaluate: `int_1^2 x/(1 + x^2)  "d"x`


Evaluate: `int_0^(pi/2) cos^3x  "d"x`


Evaluate: `int_0^(pi/4) (tan^3x)/(1 + cos 2x)  "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_(-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_0^3 x^2 (3 - x)^(5/2)  "d"x`


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


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)  (cos2x)/(1 + cos 2x + sin 2x)  "d"x`


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


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


Evaluate `int_(π/6)^(π/3) cos^2x  dx`


The value of `int_2^(π/2) sin^3x  dx` = ______.


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


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


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×