Advertisements
Advertisements
प्रश्न
Evaluate: `int_0^1(x + 1)^2 "d"x`
Advertisements
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^oo xe^-x.dx`
`int_(pi/5)^((3pi)/10) sinx/(sinx + cosx) "d"x` =
`int_0^1 (x^2 - 2)/(x^2 + 1) "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^(pi/2) (sin2x)/(1 + sin^2x) "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_0^(pi/2) (sin^2x)/(1 + cos x)^2 "d"x`
Evaluate: `int_(-1)^1 |5x - 3| "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/4) sec^4x "d"x`
Evaluate: `int_0^(pi/2) cos x/((1 + sinx)(2 + sinx)) "d"x`
Evaluate: `int_(-1)^1 1/("a"^2"e"^x + "b"^2"e"^(-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^(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 1/(3 + 2sinx + cosx) "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_0^(π/2) sin^8x dx`
Evaluate:
`int_(-π/2)^(π/2) |sinx|dx`
`int_0^1 x^2/(1 + x^2)dx` = ______.
Find the value of ‘a’ if `int_2^a (x + 1)dx = 7/2`
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`
If `int_0^π f(sinx)dx = kint_0^π f(sinx)dx`, then find the value of k.
