English

If dk∫01dx1+x-x=k3, then k is equal to ______. - Mathematics and Statistics

Advertisements
Advertisements

Question

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

Options

  • `sqrt(2)(2sqrt(2) - 2)`

  • `sqrt(2)/3(2 - 2sqrt(2))`

  • `(2sqrt(2) - 2)/3`

  • `4sqrt(2)`

MCQ
Fill in the Blanks
Advertisements

Solution

If `int_0^1 ("d"x)/(sqrt(1 + x) - sqrt(x)) = "k"/3`, then k is equal to `bb(underline(4sqrt(2))`.

Explanation:

Step 1: Rationalizing the Denominator

`sqrt(1+x)-sqrtx = ((1+x)-x)/(sqrt(1+x) + sqrtx) = 1/(sqrt(1+x) + sqrtx)`

`I = int_0^1(sqrt(1+x) + sqrtx)dx`

Step 2: Evaluating Each Integral

`I = int_0^1 sqrt(1+x)  dx + int_0^1sqrtx  dx`

Using standard integration formulas:

`(4sqrt2 - 2)/3`

`2/3`

`I = (4sqrt2 - 2)/3 + 2/3 = (4sqrt2)/3`

Step 3: Finding k

`I = k/3`

`k = 4sqrt2`

shaalaa.com
Methods of Evaluation and Properties of Definite Integral
  Is there an error in this question or solution?
Chapter 2.4: Definite Integration - MCQ

RELATED QUESTIONS

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


`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` =


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


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


Evaluate: `int_0^(pi/4) sec^2 x  "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_0^1(x + 1)^2  "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/2) cos^3x  dx`


Evaluate: `int_0^pi cos^2 x  "d"x`


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


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


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_0^1 (log(x + 1))/(x^2 + 1)  "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)  (cos2x)/(1 + cos 2x + sin 2x)  "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_-4^5 |x + 3|dx`


Evaluate:

`int_0^(π/2) sinx/(1 + cosx)^3 dx`


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×