00:04
The definite integral from 0 to 4 of the function, square root of 16 minus x squared, will equal the area of the region that's above the x -axis minus the area of the region below the x -axis.
00:17
So let's look at the graph of this function from 0 to 4.
00:21
Here's the graph of the function from 0 to 4.
00:24
Now, you can see that this region, the function stays above the x -axis, so the function stays positive.
00:31
So the definite integral is going to equal the area of this region, the region between the curve and the x -axis.
00:41
Because this region is above the x -axis, the definite integral will equal the area of this region.
00:48
Now, what is the area of this region? how would we describe the shape of this region? it looks like the quarter of the circle.
00:55
Okay, and that's exactly what this is.
00:57
This is the quarter of the circle.
00:59
What would the radius of the circle be? well, if the circle continued, it'd go all the way around like this.
01:07
The center would be right here at the origin.
01:10
And so the radius of the circle would be four units.
01:13
One, two, three, four.
01:15
And so we want to find the area of a quarter circle when the radius is four...