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### Course: AP®︎ Calculus AB (2017 edition) > Unit 8

Lesson 3: Midpoint & trapezoidal sums# Midpoint sums

Approximating area under a curve using rectangle where the heights are the value of the function at the midpoint of each interval.

## Want to join the conversation?

- Which type is the most accurate. Is it the midpoint sum or the left or right Riemann sum?(20 votes)
- Midpoint is better than the other two as it somehow comes in btwn, so your approx is neither too low, nor too high, hope it helped :)(35 votes)

- At1:04,3:28, and4:19, why do the rectangle heights change?(4 votes)
- The heights are changing as different approximation methods are being used. At1:04midpoint formula is being used, then left Riemann sum approximation, and finally right Riemann sum approximation (respectively). The instructor is demonstrating how to solve with each different kind of approximation.(9 votes)

- In this video, how is Sal getting the height of each of his rectangles?(4 votes)
- Hi @win148! For the midpoint example, Sal is getting the height of the rectangles by using the midpoint. So for example, the midpoint between x=-1 and x=0 is x=-0.5, so the height of the rectangle whose endpoints are x=-1 and x=0 is f(-0.5) = 1.25.(5 votes)

- At1:01, Sal defines the height of all three rectangles- the first two's height as 5/4 and the third ones as 13/4. What login is behind taking such heights to find the area by mid point method?(3 votes)
- Let one of these rectangles have its left endpoint at 𝑥 = 𝑎 and its right endpoint at 𝑥 = 𝑏, which means that its midpoint will be at 𝑥 = (𝑎 + 𝑏)∕2.

Within the interval [𝑎, 𝑏] it is much more common for a function 𝑓(𝑥) to be strictly increasing/decreasing rather than not, which means that its lowest and highest values within that interval will most likely be 𝑓(𝑎) and 𝑓(𝑏), while 𝑓((𝑎 + 𝑏)∕2) is somewhere in between.

Therefore, a midpoint sum is more or less guaranteed to be a better approximation of the area under the curve than a left- or righthand sum.(4 votes)

- for midpoint, do you multiply the heights of each rectangle by the width of each rectangle? or did he not do that because the question states all the rectangles have equal widths?(4 votes)
- I hope you guys have figured it out in the last 4 years but I'll post an answer for any others who have the same questions nowadays. Usually, you do have to multiply the height of the rectangles by the widths. When the widths are all equal, you can add up the heights and then multiply by the width once, since each number is being multiplied by the same width. (ex. ab + ac + ad = a(b + c + d)). In this video, however, the widths are all equal to 1, so he doesn't need to multiply.(1 vote)

- The Fundamental Theorem of Algebra states that a polynomial will cross the x-axis n times for n equaling the degree of the highest degree term. However, x^2 + 1 does not cross the x-axis. Please explain!(1 vote)
- This is the FTA: every non-zero, single-variable, degree n polynomial with complex coefficients has, counted with multiplicity, exactly n complex roots. There are n roots, but not all roots are necessarily real, so the polynomial may not cross the x axis. In fact, the roots of x^2 + 1 are i and -i.(6 votes)

- What is the summation notation for midpoint sums?(3 votes)
- Assuming a is the start, b is the end and t is the number of rectangles in between, Sum from x=0 to t-1 of (b-a)/t*f(a+(b-a)/(2t)+x(b-a)/t)

That may be a little complicated but if you break it up (b-a)/t is the length of each rectangle and the function is the height at a certain point, so we needed to get what each midpoint would be. Also, we go to t-1 instead of t because the first rectangle's midpoint is going to be x=0, the second midpoint is x=1, and we do this all the way down to the t-th midpoint being x=t-1. Hopefully that makes sense.

To find the midpoint we want to start at a, then go to the midpoint of the first rectangle, which is half the length so plus (b-a)/t divided by 2 which leads to (b-a)/(2t) and then finally we want to add another rectangle length to get to the next midpoint, and we want to add one midpoint length over and over again for as many rectangles there are and add them up. so x*(b-a)/t.

I really really hope that made sense. Let me know if it didn't and I can try a different way of explaining.(1 vote)

- Couldn't you take the left sum, subtract the right sum, divide the difference by two, and then add that difference to the underestimation or subtract it from the overestimation to get the exact value?(1 vote)
- Not nessesarily. In some instances what you suggested may work ,but there is no guarantee that the amount that one is more than is exactly twice the value that the underestimation is less than. You would get somewhere close to the exact value but it wouldn't be the exact value(4 votes)

- Is there a way to know how to divide the area under the curve in such a way that the Riemann approximation results in the least error?(1 vote)
- To the best my knowledge no. The first step to solving this problem would be constructing the mathematical model.

So we end up the objective function minimise |riemann sum - integral f(x) dx|. Also need to define the domain of each variable.

To further make the problem more complicated there could be 100's of variable. So you end up with nonlinear optimisation with a large number of variables.

You may be interested in looking into operations research.(3 votes)

- -1^2+1=0. I am confused how 2 is(1 vote)
- You are correct -1^2 + 1 =0, however that is not the calculation the video is performing.

The calculation perform is (-1)^2 + 1 which equal 2.

What is going on with -1^2 + 1 is the exponentiation is performed first. This leaves us with -1 + 1 = 0. This is because going by order of operations unary operator i.e. negative sign is applied after exponentiation

BENDMAS:

Bracket

Exponentiation

Negative sign

Divison/Multiplication # note equal hierarchy

Addition/Subtraction # note equal hierarchy(2 votes)

## Video transcript

- [Instructor] What we
wanna 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 one. And let's think about the
area under this curve, above the x-axis, from x equals negative
one to x equals two. 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 gonna break up this interval into three equal sections that are really the bases of rectangles. And then we're gonna 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 heights of each rectangle by the value of the function at the midpoint. So we see that right over here. 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, we have split up the interval
from x equals negative one to x equals two into three equal sections, and then each of them have a width of one. If we wanted a better approximation we could do more sections
or more rectangles, but let's just see how
we would compute this. Well the width of each of these is one, 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 3/2. And so this height is going
to be negative 1/2 squared plus one. So negative 1/2 squared is 1/4 plus one, so that's 5/4. So the height here is 5/4. So you take 5/4 times one. This area is 5/4, let me write that down. So if we're doing the midpoint to define the height of each rectangle, this first one has an area of 5/4. Do it in a color you
can see, five over four. The second one, same idea, 1/2 squared plus one is 5/4 times a width of one. So 5/4 there. So let me add that. Plus 5/4. And then this third rectangle, what's its height? Well we're gonna take the
height at the midpoint, so 3/2 squared is 9/4 plus one, which is the same thing as 13/4. So it has a height of 13/4, and then a width of one, so times one, which would just give us 13/4. So plus 13/4, which would give us 23 over four which is the same thing as 5 3/4. 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 wanna do it just for kicks here, let's just do that really fast. So if we wanna look at the left endpoints of our interval, well here our left endpoint is negative one, negative one squared plus one is two, two times one gives us two. And then here the left
part of this interval is x equals zero, zero squared plus one is one, one times one is one. And now here our left endpoint is one, one squared plus one is equal to two, times one, our base, is equal to two. So here we have a situation where we take our left endpoints, where it is equal to two plus one plus two or five. Well we can also look
at the right endpoints of our intervals. So this first rectangle here, clearly under approximating the area over this first interval. Its right endpoint is zero, zero squared plus one is one, so a height of one, width
of one, has an area of one. Second rectangle here, it has a height of, look
at our right endpoint, one squared plus one is two, times our width of one, well
that's just gonna give us two. And then here our right endpoint is two, squared plus one is five,
times our width of one, gives us five. So in this case we get, when we look at our right
endpoints of our intervals, we get one plus two plus five is equal to eight. 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's
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
one to x equals two, we would get better and
better approximations of the true area.