Calculating net area using geometry
About this video
Sal evaluates the integral of g(x) from -9 to 8 using geometry. The semicircle of radius 4 above the x-axis contributes +8pi, and the regions below the axis (triangles of area 8 and 3/2 and a rectangle of area 12) are subtracted, giving 8pi - 8 - 3/2 - 12, about 3.633.
You'll be able to
- split the region between a graph and the x-axis into simple shapes.
- add areas above the x-axis and subtract areas below it to find a net area.
- evaluate an expression with pi and round to three decimal places.
Key words
- Net area
- Area above the x-axis minus area below it; this is what a definite integral gives.
- Semicircle
- Half a circle; area (1/2) pi r^2.
- Definite integral
- The signed area between a graph and the x-axis over an interval.
Transcript
- [Narrator] All right, so we want to evaluate this integral right here, and they say, "Express your answer to three decimal places." So pause this video, have a go at this before we do this together. All right, so it's a definite integral of g of x, dx from negative 9 to 8.
So we know what g of x looks like, so we're going from x equals negative 9. So it's going to be from, let me just draw that, from x equals negative 9, all the way to x equals 8. So what we're going to be thinking about is the area between this curve and the X-axis.
Now we want to be careful here, because we can absolutely figure out this area here, this area here, and then this area down here. But we're not always going to have positive values. Another way to think about it is, it is the area that is below g of x, and above the X-axis.
So for example, this part of the integral is going to be positive. but when g of x is below the X-axis, we're actually going to subtract that area. So let's calculate each of these. So this positive area here, this is a semicircle, let's see, it has a radius of, looks like its center is right over here, and it has a radius of 4.
It has a radius of 4. So the whole circle would be pi r squared, which would be 16 pi. But we don't have the whole circle there. We only have half of a circle. We have a semicircle. So we are going to divide that by 2, which is going to be equal to 8 pi. That's the area for just this circle.
And that's gonna be positive. But then we're going to want subtract these areas here. So this triangle right over here, if we do base times height times one half, let's see, the base looks like this is 1, 2, 3, 4. Height is also 4. So if we do 4 times 4 is 16 times 1/2 is 8.
So that's going to be 8 over there. And then this area, actually, let's just break it up into two pieces. We have the small triangle that is 1 times 3 times 1/2. So 1 times 3 times 1/2 is three halves. That area there is three halves. And then this area over here, this rectangle, is 3 times 4.
And so that is going to be 12. Now remember what I said, the positive areas when we are below g of x and above the X-axis. So that's going to be our 16 pi, or sorry, our 8 pi. So that is going to be our 8 pi there. And then all of these areas we're going to subtract when we're evaluating this definite integral, because it is above our g of x and below the X axis, it's gonna be 8 pi minus 8, minus three halves, minus 12.
And I can get a calculator out for this. So I am going to get 8 times. And then we have pie right over here. And then from that I'm going to subtract 8. Then I'm going to subtract three halves. So minus three halves, same thing as 1.5. And then I'm going to subtract 12.
And I'm going to get, if I want to round the answer to three decimal places, I get 3.633. So this is approximately, approximately 3.633. And we're done. Square units. And remember, this part was positive up here. And then we subtract these parts right down here.
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