Advertisements
Advertisements
प्रश्न
Evaluate `int_-1^1 |x^4 - x|dx`.
उत्तर
Let I = `int_-1^1 |x^4 - x|dx`
= `int_-1^0 (x^4 - x)dx - int_0^1 (x^4 - x)dx`
= `[x^5/5 - x^2/2]_-1^0 - [x^5/5 - x^2/2]_0^1`
= `[(0 - 0) - ((-1)/5 - 1/2)] - [(1/5 - 1/2) - 0]`
= `7/10 + 3/10`
= 1.
APPEARS IN
संबंधित प्रश्न
Evaluate the definite integrals `int_0^pi (x tan x)/(sec x + tan x)dx`
Evaluate : `int "e"^(3"x")/("e"^(3"x") + 1)` dx
`int_2^7 sqrt(x)/(sqrt(x) + sqrt(9 - x)) dx` = ______.
Evaluate `int_1^2 (sqrt(x))/(sqrt(3 - x) + sqrt(x)) "d"x`
`int_0^(pi/4) (sec^2 x)/((1 + tan x)(2 + tan x))`dx = ?
`int_0^1 (1 - x/(1!) + x^2/(2!) - x^3/(3!) + ... "upto" ∞)` e2x dx = ?
`int_0^(pi"/"4)` log(1 + tanθ) dθ = ______
`int_2^3 x/(x^2 - 1)` dx = ______
`int_"a"^"b" sqrtx/(sqrtx + sqrt("a" + "b" - x)) "dx"` = ______.
`int_0^{pi/2} cos^2x dx` = ______
`int_{pi/6}^{pi/3} sin^2x dx` = ______
`int_0^1 "dx"/(sqrt(1 + x) - sqrtx)` = ?
If `int_0^"k" "dx"/(2 + 32x^2) = pi/32,` then the value of k is ______.
`int_0^9 1/(1 + sqrtx)` dx = ______
`int_(-pi/4)^(pi/4) 1/(1 - sinx) "d"x` = ______.
`int_(-1)^1 (x^3 + |x| + 1)/(x^2 + 2|x| + 1) "d"x` is equal to ______.
`int_0^(pi/2) (sin^"n" x"d"x)/(sin^"n" x + cos^"n" x)` = ______.
`int_0^(pi/2) cos x "e"^(sinx) "d"x` is equal to ______.
`int_a^b f(x)dx` = ______.
`int_0^1|3x - 1|dx` equals ______.
Let `int_0^∞ (t^4dt)/(1 + t^2)^6 = (3π)/(64k)` then k is equal to ______.
`int_0^(π/4) x. sec^2 x dx` = ______.
If `int_0^(2π) cos^2 x dx = k int_0^(π/2) cos^2 x dx`, then the value of k is ______.
`int_-1^1 |x - 2|/(x - 2) dx`, x ≠ 2 is equal to ______.
If `int_0^1(3x^2 + 2x+a)dx = 0,` then a = ______
Evaluate the following integral:
`int_-9^9 x^3/(4 - x^2) dx`
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 definite intergral:
`int_1^3logx dx`
