मराठी

Evaluate ∫2−2 x2/(1+5x) dx

Advertisements
Advertisements

प्रश्न

 
 

Evaluate `int_(-2)^2x^2/(1+5^x)dx`

 
 
Advertisements

उत्तर

 

 Consider the given integral

`I= int_(-2)^2x^2/(1+5^x)dx`

Let us use the property,

`int_a^bf(x)dx=int_b^af(a+b-x)dx`

`:.I = int_(-2)^2(-x)^2/(1+5^(-x))dx`

 `=int_(-2)^2(5^(x)x^2)/(1+5^x)dx `

 Adding equations (1) and (2), we have,

`2I=int_(-2)^2(1+5^x)/(1+5^x)xx x^2dx`

`=int_(-2)^2x^2dx`

`=[x^3/3]^2`

`=1/3[8-(8)]`

`=1/3[16]`

`=>I= 8/3`

 

 
shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2015-2016 (March) All India Set 1 N

संबंधित प्रश्‍न

If `int_0^alpha3x^2dx=8` then the value of α is :

(a) 0

(b) -2

(c) 2 

(d) ±2


By using the properties of the definite integral, evaluate the integral:

`int_0^(pi/4) log (1+ tan x) dx`


By using the properties of the definite integral, evaluate the integral:

`int_(pi/2)^(pi/2) sin^7 x dx`


If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that

\[\int_a^b xf\left( x \right)dx = \left( \frac{a + b}{2} \right) \int_a^b f\left( x \right)dx\]

Evaluate : `int _0^(pi/2) "sin"^ 2  "x"  "dx"`


Choose the correct alternative:

`int_(-9)^9 x^3/(4 - x^2)  "d"x` =


Evaluate `int_1^2 (sqrt(x))/(sqrt(3 - x) + sqrt(x))  "d"x`


`int_0^{pi/2}((3sqrtsecx)/(3sqrtsecx + 3sqrt(cosecx)))dx` = ______ 


`int_0^{pi/2} xsinx dx` = ______


`int_0^{pi/4} (sin2x)/(sin^4x + cos^4x)dx` = ____________


If f(x) = |x - 2|, then `int_-2^3 f(x) dx` is ______


`int_0^1 "dx"/(sqrt(1 + x) - sqrtx)` = ?


`int_0^(pi/2) 1/(1 + cos^3x) "d"x` = ______.


Show that `int_0^(pi/2) (sin^2x)/(sinx + cosx) = 1/sqrt(2) log (sqrt(2) + 1)`


`int_(-2)^2 |x cos pix| "d"x` is equal to ______.


Evaluate the following:

`int_(-pi/4)^(pi/4) log|sinx + cosx|"d"x`


The value of the integral `int_(-1)^1log_e(sqrt(1 - x) + sqrt(1 + x))dx` is equal to ______.


If `β + 2int_0^1x^2e^(-x^2)dx = int_0^1e^(-x^2)dx`, then the value of β is ______.


What is `int_0^(π/2)` sin 2x ℓ n (cot x) dx equal to ?


If `int_0^(2π) cos^2 x  dx = k int_0^(π/2) cos^2 x  dx`, then the value of k is ______.


Evaluate the following definite integral:

`int_4^9 1/sqrt"x" "dx"`


Evaluate the following definite integral:

`int_-2^3 1/(x + 5) dx`


Evaluate: `int_-1^1 x^17.cos^4x  dx`


Solve the following.

`int_0^1 e^(x^2) x^3dx`


Evaluate the following integral:

`int_0^1 x(1-x)^5 dx`


Evaluate the following integral:

`int_-9^9x^3/(4-x^2)dx`


Evaluate the following integral:

`int_0^1x(1-x)^5dx`


Solve the following.

`int_0^1e^(x^2)x^3dx`


The value of \[\int_{-1}^{1}\left(\sqrt{1+x+x^{2}}-\sqrt{1-x+x^{2}}\right)\mathrm{d}x\] is


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×