English

Let I1 = ∫ee2 1logx dx and I2 = ∫12exx dx then - Mathematics and Statistics

Advertisements
Advertisements

Question

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

Options

  • I1 = `1/3 "I"_2`

  • I1 + I2 = 0 

  • I1 = 2I2 

  • I1 = I2 

MCQ
Advertisements

Solution

I1 = I2 

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/4) cot^2x.dx`


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


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


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


Evaluate the following:

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


Choose the correct option from the given alternatives : 

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


`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_0^1 |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_0^1(x + 1)^2  "d"x`


Evaluate: `int_(pi/6)^(pi/3) sin^2 x  "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^(pi/2) (sin^4x)/(sin^4x + cos^4x)  "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 1/sqrt(3 + 2x - x^2)  "d"x`


Evaluate: `int_0^1 x* tan^-1x  "d"x`


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


Evaluate:

`int_(π/4)^(π/2) cot^2x  dx`.


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


Evaluate:

`int_-4^5 |x + 3|dx`


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


Evaluate:

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