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Choose the correct alternative: ∫-99x34-x2 dx = - Mathematics and Statistics

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प्रश्न

Choose the correct alternative:

`int_(-9)^9 x^3/(4 - x^2)  "d"x` =

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अध्याय 1.6: Definite Integration - Q.1

संबंधित प्रश्न

Evaluate :`int_0^pi(xsinx)/(1+sinx)dx`


Evaluate : `intsec^nxtanxdx`


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

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


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


Evaluate : `int  "e"^(3"x")/("e"^(3"x") + 1)` dx


Evaluate :  `int 1/sqrt("x"^2 - 4"x" + 2) "dx"`


`int_2^7 sqrt(x)/(sqrt(x) + sqrt(9 - x))  dx` = ______.


Evaluate `int_1^3 x^2*log x  "d"x`


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


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


`int_"a"^"b" sqrtx/(sqrtx + sqrt("a" + "b" - x)) "dx"` = ______.


If `int_0^"a" sqrt("a - x"/x) "dx" = "K"/2`, then K = ______.


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


f(x) =  `{:{(x^3/k;       0 ≤ x ≤ 2), (0;     "otherwise"):}` is a p.d.f. of X. The value of k is ______


`int_0^{pi/2} cos^2x  dx` = ______ 


`int_-2^1 dx/(x^2 + 4x + 13)` = ______


`int_0^pi sin^2x.cos^2x  dx` = ______ 


`int_0^1 log(1/x - 1) "dx"` = ______.


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Find `int_2^8 sqrt(10 - x)/(sqrt(x) + sqrt(10 - x)) "d"x`


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


If `int_0^"a" 1/(1 + 4x^2) "d"x = pi/8`, then a = ______.


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The value of `int_0^1 tan^-1 ((2x - 1)/(1 + x - x^2))  dx` is


Evaluate: `int_(pi/6)^(pi/3) (dx)/(1 + sqrt(tanx)`


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The value of the integral `int_0^sqrt(2)([sqrt(2 - x^2)] + 2x)dx` (where [.] denotes greatest integer function) is ______.


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 limit :

`lim_("x"->3)[sqrt("x"+6)/"x"]`


`int_1^2 x logx  dx`= ______


`int_0^(2a)f(x)/(f(x)+f(2a-x))  dx` = ______


Evaluate `int_1^2(x+3)/(x(x+2))  dx`


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Evaluate the following integral:

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Evaluate the following definite intergral:

`int_1^3logx  dx`


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