Advertisements
Advertisements
प्रश्न
Evaluate: `int_0^1(x + 1)^2 "d"x`
उत्तर
`int_0^1(x + 1)^2 "d"x = [(x + 1)^3/3]_0^1`
= `1/3[(1 + 1)^3 - (0 + 1)^3]`
= `1/3(8 -1)`
= `7/3`
APPEARS IN
संबंधित प्रश्न
Evaluate: `int_0^(π/4) cot^2x.dx`
Evaluate: `int_0^1 (x^2 - 2)/(x^2 + 1).dx`
Evaluate: `int_0^(pi/2) x sin x.dx`
Evaluate: `int_0^oo xe^-x.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_0^(x/4) sqrt(1 + sin 2x) "d"x` =
`int_(pi/5)^((3pi)/10) sinx/(sinx + cosx) "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_0^1 |x| "d"x`
Evaluate: `int_0^1 1/sqrt(1 - x^2) "d"x`
Evaluate: `int_0^pi cos^2 x "d"x`
Evaluate: `int_0^9 sqrt(x)/(sqrt(x) + sqrt(9 - x) "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) 1/(5 + 4cos x) "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^pi x*sinx*cos^2x* "d"x`
Evaluate: `int_(-1)^1 (1 + x^2)/(9 - x^2) "d"x`
Evaluate: `int_0^(pi/4) log(1 + tanx) "d"x`
Evaluate: `int_0^pi 1/(3 + 2sinx + cosx) "d"x`
Evaluate: `int_0^(π/4) sec^4 x dx`
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_(-π/2)^(π/2) |sinx|dx`
Evaluate `int_(π/6)^(π/3) cos^2x dx`
Evaluate:
`int_-4^5 |x + 3|dx`
The value of `int_2^(π/2) sin^3x dx` = ______.
Evaluate:
`int_0^(π/2) sinx/(1 + cosx)^3 dx`
Prove that: `int_0^1 logx/sqrt(1 - x^2)dx = π/2 log(1/2)`
Evaluate `int_(-π/2)^(π/2) sinx/(1 + cos^2x)dx`