What is the anti-derivitive of |x|?

Answer 1

∫|x|dx = |x²|/2 + C = x²/2 + C

Since x² is always positive.

Answer 2

The antiderivative splits into two parts:

|x| = x if x > 0 or -x if x<0

So the antiderivative would be

0.5 x^2 if x>0

-0.5 x^2 if x<0

Answer 3

x^2

Answer 4

Consider the differential equation dy/dx=xn (n does not equal -1). What is the antiderivative of this equation - ie, what is y in terms of x? Consider
y=1/(n+1)*xn+1 - what is the derivative of y with respect to x? It is y'=xn! We express this result in the following form.
dy/dx=xn
dy=xndx .............multiply both sides by dx
[inte] dy= [inte] xndx ..........integrate both sides
y=1/(n+1)*xn+1 + C
The integral sign merely tells you to find the antiderivative of the equation. The "dx" and "dy" tell you what variable you are integrating (antidifferentiating) with respect to. To the left of the "dy" and "dx" is the derivative you are trying to undo.
So the left hand side [inte] dy = [inte] 1*dy means what function of y has a first derivative (taken wrt y) equal to 1? Obviously, it is f(y)=y since df/dy=dy/dy=1. The right hand side means what function of x has a derivative of xn? This is the solution to the integral.

Answer 5

x>=0 then , |x|=x!

so S(|x|dx)=S(xdx)=x^2/2
x<0 the |x|=-x
so S(|x|dx)=S(-xdx)=-x^2/2

solution: x>=0, then x^2/2 and x<0 then -x^2/2

Answer 6

y

Answer 7

.5x^2