00:01
Suppose we have g of x is equal to the integral from 0 to x of f of t dt, and that we are given the graph of f of t.
00:09
For the first part, we want to find the value of g at x equals 0, 1, 2, 3, and 6.
00:17
So for g of 0, we have integral from 0 to 0 of f of t, and because the upper and lower limits are the same, then this integral here equals 0 immediately.
00:34
And then we have g of 1, which equals the integral from 0 to 1 of f of t, and that's the area of the region bounded by f of t, and the t -axis, so it should be this area here that is from 0 to 1, and that's the area of the rectangle, that's length times the width, length equals 2, times the width of 1, that equals 2.
01:05
And then you have g of 2.
01:07
G of 2 is the integral from 0 to 2 of f, and that's going to be g of 1 plus the area of the region from 1 to 2, which is a trapezoid, so it'll be 1 half times upper base plus lower base, upper base is 2, lower base is going to be 4, and we multiply this by the height, which is 1.
01:39
So this equals 2 plus 1 half of 6, that's 3, so this equals 5.
01:47
And then you have g of 3.
01:49
G of 3 equals the integral from 0 to 3 of f of t dt, and that's going to be the area of the region over the interval 0 to 3, so it'll be from 0 to 2 plus from 2 to 3.
02:06
So the area from 0 to 2 is already there, that's g of 2, and we're going to add this to the area of the region that is on the interval 2 to 3.
02:18
If you can see that this is a triangle, then that's 1 half base times the height, base there is 1, height there is 4, so it'll be 5 plus 1 half of 4, which is 2, that's equals 7.
02:33
And then lastly we have g of 6.
02:36
G of 6 is the integral from 0 to 3, or 0 to 6 of f of t dt, which we can separate into integral from 0 to 3 plus the integral from 3 to 6 of f of t dt.
02:59
So the first integral that's actually g of 3, which is 7, plus the integral from 3 to 6 will be this area here, which again is a trapezoid.
03:12
So with that trapezoid, we have 1 half times upper base, which is 1, plus its lower base, which is 3, times its height, which is negative 2.
03:26
So it'll be 7 plus half of 4, which is 2, times negative 2, that's negative 4.
03:39
So 7 minus 4, that equals 3.
03:44
And then we are asked where g is increasing.
03:48
So for part b, our g is increasing whenever the derivative of g is greater than 0.
03:57
Now g prime of x, by the fundamental theorem of calculus, that's a derivative with respect to x, so the integral from 0 to x of f, and that equals f of x.
04:10
So we want to find that interval where f of x is greater than 0...