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Main content
Current time:0:00Total duration:5:29
AP.CALC:
LIM‑5 (EU)
,
LIM‑5.A (LO)
,
LIM‑5.A.1 (EK)
,
LIM‑5.A.2 (EK)
,
LIM‑5.A.3 (EK)
,
LIM‑5.A.4 (EK)

Video transcript

what we want to do in this video is get an understanding of how we can approximate the area under a curve and for the sake of an example we'll use the curve y is equal to x squared plus 1 and let's think about the area under this curve above the x-axis from x equals negative 1 to x equals 2 so that would be this area right over here and there's many ways that I could tackle this but what I'm going to do is I'm going to break up this interval into three equal sections that are really the basis of rectangles and we're going to think about the different ways to define the heights of those rectangles so once again I'm going to approximate using three rectangles of equal width and then we'll think about the different ways that we can define the heights of the rectangles so let's first define the height of each rectangle by the value of the function at the midpoint so we see that right over here and so let's just make sure that it actually makes sense to us so if we look at our first rectangle right over here actually let's just first appreciate we have split up this X that we've split up the interval from x equals negative 1 to x equals 2 into 3 equal sections and then each of them have a width of 1 if we wanted a better approximation we could do more sections or more rectangles also see how we would compute this well the width of each of these is 1 the height is based on the value of the function at the midpoint the midpoint here is negative 1/2 the midpoint here is 1/2 the midpoint here is three-halves and so this height is going to be negative 1/2 squared plus 1 so negative 1/2 squared is 1/4 plus 1 so that's 5/4 so the height here is 5/4 so you take 5/4 times 1 this area is 5/4 let me write that down so for doing the midpoint to define the height of each rectangle this first one has an area of 5/4 I'm doing a color you can see five over for the second one same idea 1/2 squared plus one is five fourths times the width of one so five fourths there so let me add that plus 5/4 and then this third rectangle what's it tight well we're going to take the height of the midpoint so three halves squared is nine fourths plus one which is the same thing as thirteen fourths so it has a height of thirteen fourth and then a width of 1 so times 1 which would just give us thirteen fourths so plus thirteen fourths which would give us 23 over four which is the same thing as five and three fourths and so this is often known as a midpoint approximation where we're using the midpoint of each interval to define the height of our rectangle but this isn't the only way to do it we could look at the left endpoint or the right endpoint and we do that in other videos and if we want to do it just for kicks here let's just do that really fast so if we want to look at the left endpoint of our interval well here our left endpoint is negative 1 negative 1 squared plus 1 is 2 2 times 1 gives us 2 and then here the left part of this interval is x equals 0 0 squared plus 1 is 1 1 times 1 is 1 and now here our left endpoint is 1 1 squared plus 1 is equal to 2 times 1 our base is equal to 2 so here we have a situation where we take our left endpoints where it is equal to 2 plus 1 plus 2 or 5 but we can also look at the right endpoint of our intervals so this first rectangle here clearly under approximating the area over this first interval its right endpoint is 0 0 squared plus 1 is so height of one width of one has an area of one second rectangle here it has a height of we look at our right endpoint 1 squared plus 1 is 2 times our width of 1 well that's just going to give us 2 and then here our right endpoint is 2 squared plus 1 is 5 times our width of 1 gives us 5 so in this case we get when we look at our right endpoints of our intervals we get 1 plus 2 plus 5 is equal to 8 and eyeballing this it looks like we're definitely over counting more than under counting and so this looks like an over approximation so the whole idea here is just to appreciate how we can compute these approximations using rectangles and as you can imagine if we added more rectangles that had skinnier and skinnier bases but still covered the interval from x equals negative 1 to x equals 2 we would get better and better approximations of the true area
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