00:01
Okay, so for this function, they want you to take the derivative of this integral function and using the second part of the fundamental theorem of calculus.
00:11
And the second part of the fundamental theorem of calculus states that if we want to take the derivative from some constant a to x of some function, f of t, in respect to t, then that derivative would just be f of x.
00:25
All right, so since we're dealing with an absolute value of a function, you have to keep in mind of the zeros or when the function crosses the x -axis.
00:36
And conveniently for this problem, it's when x equals to zero, because for this function, the absolute value of x looks like this v function.
00:46
So since we're integrating just one side of the function, either from the left or to the right, depending on whether u is positive or negative, you can just consider just integrating half or treating it as one function.
01:04
So if you was, so if you was either, since we're dealing with only either positive values or negative values, we consider as u equal to, if you was greater than zero, the absolute value of x would equal to x.
01:22
And if you was less than zero, then the absolute value of x would equal to negative x.
01:28
And if we were to apply our fundamental theory of calculus to a new, a new derivative of an integral function from you 0 to u of x d x, it would equal to u...