हिंदी

Evaluate: ∫03x2(3-x)52 dx - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Evaluate: `int_0^3 x^2 (3 - x)^(5/2)  "d"x`

योग

उत्तर

Let I = `int_0^3 x^2 (3 - x)^(5/2)  "d"x`

= `int_0^3(3 - x)^2[3 - (3 - x)]^(5/2)  "d"x`   ....`[∵ int_0^"a" "f"(x) "d"x = int_0^"a" "f"("a" - x)"d"x]`

= `int_0^3(9 - 6x + x^2)x^(5/2)  "d"x`

= `int_0^3(9x^(5/2) - 6x^(7/2) + x^(9/2)) "d"x`

= `9int_0^2 x^(5/2) "d"x - 6 int_0^3 x^(7/2)"d"x + int_0^3 x^(9/2) "d"x`

= `9[(x^(7/2))/(7/2)]_0^3 - 6[(x^(9/2))/(9/2)]_0^3 + [(x^(11/2))/(11/2)]_0^3`

= `18/7[(3)^(7/2) - 0] - 12/9[(3)^(9/2) - 0] + 2/1[(3)^(11/2) - 0]`

= `[18/7 - (12/9 xx3) + (2/11 xx 9)](3)^(7/2)`

= `((198 - 308 + 126)/77)(3)^(7/2)`

∴ I = `16/77(3)^(7/2)`

shaalaa.com
Methods of Evaluation and Properties of Definite Integral
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2.4: Definite Integration - Short Answers II

संबंधित प्रश्न

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


Evaluate: `int_0^(pi/2) x sin x.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_0^(pi/2) log(tanx)  "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^(pi/4) sec^2 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_(pi/6)^(pi/3) sin^2 x  "d"x`


Evaluate: `int_0^(pi/4)  cosx/(4 - sin^2 x)  "d"x`


Evaluate: `int_1^3 (cos(logx))/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/sqrt(2)) (sin^-1x)/(1 - x^2)^(3/2)  "d"x`


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


Evaluate: `int_0^"a" 1/(x + sqrt("a"^2 - x^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_(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_0^pi x*sinx*cos^2x* "d"x`


Evaluate: `int_(-1)^1 (1 + x^2)/(9 - 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`


If `int_2^e [1/logx - 1/(logx)^2].dx = a + b/log2`, then ______.


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


Evaluate:

`int_0^(π/2) sin^8x  dx`


Evaluate:

`int_0^(π/2) (sin 2x)/(1 + sin^4x)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×