It is x|x|/2 + C
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The absolute value of 19 is 19. If x is positive , absolute x equals x.
zero. The absolute value of a number is just the positive version of that number, so the absolute value of x is x, and x minus x is zero.
Antiderivative of x/-1 = -1(x^2)/2 + C = (-1/2)(x^2) + C Wolfram says antiderivative of x^-1 is log(x) + C
If f(x)=1/x then F(x)=antiderivative of f(x)=ln(|x|) (the natural log of the absolute value of x) There's another way of reading this question. The anti derivative of 1 is x+c. Dividing that by x gives you 1 + c/x
If a number is not less than zero then that is its absolute value. If a number is less than zero, its negative is its absolute value. So, if |x| denotes the absolute value of x, then |x| = -x for x<0 [since if x<0 then -x>0] and |x| = x for x>= 0