00:01
Suppose we're given this graph of the derivative of f and that f of 0 equals 3.
00:04
Now by the fundamental theorem of calculus, we know that f of x minus f of 0 equals the integral from 0 to x of f prime of t d t.
00:19
So since we're given the value of f of 0, then this is the same as f of x equal to integral from 0 to x of f prime of t d t, plus f of 0, which is 3.
00:31
So when x is equal to 1, we have f of 1 equal to the integral from 0 to 1 of f prime of t d t plus 3, this is going to be the area of the region bounded by f prime over the interval 0 to 1, which will be this rectangle here.
00:51
That'll be length times the width, that's 1 times 2, and then plus 3, that's equal to 5.
00:59
When x is equal to 2, you have f of 2 equal to the integral from 0 to 2 of f prime of t d t plus 3.
01:09
That's going to be area of the region over the interval 0 to 2.
01:14
That's 2 times 2 and then plus 3 equal to 7.
01:18
When x is equal to 3, you have f of 3 equal to the integral from 0 to 3 of f prime of t d t plus 3...