00:01
So in this question, they give us a graph of a function f of x shown above, and using the geometry of the graph, we want to evaluate each of the following definite intervals.
00:12
In a, i want the integral from 0 to 2 of f of x dx.
00:19
Remember, a definite interval computes the net -signed area between my function and the x -axis over the appropriate interval.
00:30
So on the interval from 0 to 2, i'm looking at the area of this rectangle.
00:41
Now, this rectangle has a width of 2, and it's got a height of 3, 4, 5, 6, 7, 8, 9.
00:50
So i have an area of 18 here.
00:54
However, that area lies below the x -axis.
00:58
Therefore, that definite integral is counted as negative by the definite interval, giving me negative 18.
01:09
In b, i want the interval from 2 to 4 of f of x d.
01:14
So now i'm looking at again my net side area between my function and the horizontal axis over that interval from x equals 2 to x to x equals 4.
01:29
The area of this triangle, one half base times height, is one half, my base is two, my height is nine, i'm getting nine.
01:42
However, again, that area is below the x -axis, and so that area is counted as negative by the definite interval.
01:51
My integral from 2 to 4 of f of x, dx, is negative 9.
01:57
Then in c, i want the integral from 0 to 4 of f of x d x.
02:04
To do that, i just add the integral from 0 to 2 to the integral from 2 to 4, giving me negative 27...