00:01
Consider the graph of the derivative of g below.
00:02
Let's suppose the value of g at x equals 0 is negative 4.
00:07
For the first part, we want to find the value of g at x equals 2.
00:12
Now, based on this figure and using fundamental theorem of calculus, g of 2 minus g of 0, this is equal to the integral from 0 to 2 of g prime of x d x.
00:27
So, g of 2 is equal to the integral from 0 to 2, g prime of x dx, plus g of 0.
00:37
Now, the integral from 0 to 2 of g prime of x d x is the area of the region bounded by the graph of g prime of x and the x axis over the interval 0 to 2.
00:51
So based on this figure, it should be this region.
00:55
That's equal to area of the triangle on the interval 0 to 1 plus the area of the triangle over the interval 1 to 2.
01:06
So that's 1 half base of 1 times the height of negative 1.
01:11
That's for the first triangle.
01:13
And for the upper triangle, that's 1 half times a base of 1 times the height of 1.
01:20
And then plus g of 0, which we know is negative 4...