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Finding definite integrals using area formulas

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.A (LO)
,
FUN‑6.A.1 (EK)
Since definite integrals are the net area between a curve and the x-axis, we can sometimes use geometric area formulas to find definite integrals. See how it's done.

Video transcript

- [Instructor] We're told to find the following integrals, and we're given the graph of f right over here. So this first one is the definite integral from negative six to negative two of f of x dx. Pause this video and see if you can figure this one out from this graph. All right we're going from x equals negative six to x equals negative two, and the definite integral is going to be the area below our graph and above the x-axis. So it's going to be this area right over here. And how do we figure that out? Well this is a semicircle, and we know how to find the area of a circle if we know its radius. And this circle has radius two, has a radius of two. No matter what direction we go in from the center, it has a radius of two. And so the area of a circle is pi r squared. So it would be pi times our radius which is two squared, but this is a semicircle, so I'm gonna divide by two. It's only 1/2 the area of the full circle. So this is going to be four pi over two, which is equal to two pi. All right let's do another one. So here we have the definite integral from negative two to one of f of x dx. Pause the video and see if you can figure that out. All right let's do it together. So we're going from negative two to one, and so we have to be a little bit careful here. So the definite integral, you could view it as the area below the, below the function and above the x-axis. But here the function is below the x-axis. And so what we can do is, we can figure out this area, just knowing what we know about geometry, and then we have to realize that this is going to be a negative value for the definite integral because our function is below the x-axis. So what's the area here? Well there's a couple of ways to think about it. We could split it up into a few shapes. So you could just view it as a trapezoid or you can just split it up into a rectangle and two triangles. So if you split it up like this, this triangle right over here has an area of one times two times 1/2. So this has an area of one. This rectangle right over here has an area of two times one, so it has an area of two. And then this triangle right over here is the same area as the first one. It's going to have a base of one, a height of two, so it's one times two times 1/2. Remember the area of a triangle is 1/2 base times height. So it's one. So if you add up those areas, one plus two plus one is four, and so you might be tempted to say oh is this going to be equal to four? But remember our function is below the x-axis here, and so this is going to be a negative four. All right let's do another one. So now we're gonna go from one to four of f of x dx. So pause the video and see if you can figure that out. So we're gonna go from here to here, and so it's gonna be this area right over there. So how do we figure that out? Well it's just the formula for the area of a triangle, base times height times 1/2. So or you could say 1/2 times our base, which is a length of, see we have a base of three right over here, go from one to four, so 1/2 times three times our height, which is one, two, three, four, times four. Well this is just going to get us six. All right last but not least, if we are going from four to six of f of x dx. So that's going to be this area right over here, but we have to be careful. Our function is below the x-axis, so we'll figure out this area and then it's going to be negative. So this is a half of a circle of radius one. And so the area of a circle is pi times r squared, so it's pi times one squared. That would be the area if we went all the way around like that, but this is only half of the circle, so divided by two. And since this area is above the function and below the x-axis, it's going to be negative. So this is going to be equal to negative pi over two. And we are done.