2006-11-30 08:44:48 · 7 answers · asked by km 1 in Science & Mathematics ➔ Mathematics
The anti-derivative of a function f(x) is another function F(x) so that F´(x) = f(x).
If f(x) = x^n (n<-1 or n>-1) then the anti-derivative is F(x) = (x^(n+1))/(n+1) + C ... we need to add an arbitrary constant.
In your case we have: f(x) = x^1 and F(x) = (x^2)/2 + C
2006-11-30 08:57:39 · answer #1 · answered by vahucel 6 · 1⤊ 1⤋
Just add one to the exponent and then divide by that number.
The antiderivative of x^1 is x^(1+1) / (1+1) + C
= x^2 / 2 + C
= (1/2)x^2 + C
2006-11-30 16:45:46 · answer #2 · answered by MsMath 7 · 0⤊ 1⤋
the antiderivative of x^p = [x^(p+1)] / (p+1)
2006-11-30 16:47:30 · answer #3 · answered by Anonymous · 0⤊ 1⤋
(1/2)x^2
2006-11-30 16:46:55 · answer #4 · answered by Josh H 2 · 0⤊ 1⤋
(1/2)x^2
2006-11-30 16:46:26 · answer #5 · answered by Herb J 2 · 0⤊ 1⤋
Add one to the exponent and then divide by that number.
(1/2)x^2
2006-11-30 16:47:07 · answer #6 · answered by barnacle1988 3 · 0⤊ 0⤋
Y
I mean why do you want to know
2006-11-30 16:47:18 · answer #7 · answered by Anonymous · 0⤊ 1⤋