Main content

### Course: Integral Calculus > Unit 1

Lesson 4: Riemann sums in summation notation- Riemann sums in summation notation
- Riemann sums in summation notation
- Worked example: Riemann sums in summation notation
- Riemann sums in summation notation
- Midpoint and trapezoidal sums in summation notation
- Riemann sums in summation notation: challenge problem

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Midpoint and trapezoidal sums in summation notation

Estimating the area under a curve with trapazoids instead of rectangles can give a closer approximation. Created by Sal Khan.

## Want to join the conversation?

- Isn't the formula for estimating area using the midpoint, and for using trapezoids the same?(46 votes)
- the difference is that the midpoint method takes the average of the two x values, while the trapezoid method takes the average of the two y values(56 votes)

- Which of those is the most accurate?(4 votes)
- Some times it depends on the function being modeled, but in general, the trapezoid rule is most accurate when Δx is larger. As Δx becomes smaller and smaller, to the infinitesimally small dx, all the different forms converge to the same value.(17 votes)

- Which method gives you the most accurate approximation?(10 votes)
- For a fixed value of
`n`

, none of the approximations is best for all functions. For example, if you give me`n`

, I'll say approximate

1+cos(πx)

between`0`

and`2n`

. That puts the rectangle boundaries at the even numbers between`0`

and`2n`

, and for any even integer`x`

,`1+cos(πx)`

is`1+1`

, or`2`

. So the left, right, and trapezoidal approximations all look like`n`

rectangles of base 2 and height 2, for a total of`4n`

. The midpoints of all the boundaries are all the odd numbers between`0`

and`2n`

, and for any odd integer`x`

,`1+cos(πx)`

is`1-1`

, or`0`

. So the midpoint approximation is`n`

rectangles of base 2 and height 0, for a total of`0`

. All of these approximations are pretty terrible. (The correct value is`2n`

.) We have rigged the function so that all the boundary and midpoint values are at extreme values. For fixed`n`

, you can always concoct a function that will make any of the approximations look very good or very bad.

By the way, this example shows why Jazon's claim that "the average of the left and right approximations is exact" cannot be correct. The average here is`4n`

, which is far from exact.(6 votes)

- For left riemann sums, could you not also have the subscript i=1 be i=0? Seems it would be slightly clearer than having the sum begin at xi-1.(4 votes)
- That's purely a matter of choice in how you use the notation. Some people find it more confusing to begin counting from 0, which means subscripts don't match normal counting (for example, the fourth item is x sub3, not x sub4.(4 votes)

- I don't understand why we call it "x to the n - 1"

Why is it -1? Can someone please explain it to me? I never understood this.(2 votes)- Because we define i=1 under the sigma sign, but we want to evaluate the rectangle starting at x sub 0 not x sub 1, so we say x sub n-1.(7 votes)

- Under what circumstances does the sum of the trapezoids approach the definite integral of the function? How many partitions are needed?(3 votes)
- The sum of the trapezoids approaches the value of the definite integral of the function as the number of partitions approaches infinity. This is true for any of the Riemann sums and is the basis for the definition of the definite integral.(2 votes)

- Isn't taking the definite integral still more accurate? How are these techniques useful?(1 vote)
- Those two principles exist for separate reasons. The trapezoidal riemann sum (as well as LRAM, MRAM, and RRAM) are just approximations of area, and as you said, they are all less accurate than a definite integral. However, they do prove the existence of a definite integral because as the number of intervals of a rieman sum increases to infinity (provided that the bounds remain the same), the riemann sums become the integral. By fundamental theorem of calculus, the integral is the same as the antiderivative, and calculating the antiderivative is , again, far more accurate than any riemann sum.

In short, yes the definite integral is more accurate, but any riemann sum help proves the existence of an integral when the intervals increase to infinity.(5 votes)

- What does the x sub i equall or mean(2 votes)
- Hey what is we use the left boundary uptil half of the graph ie uptill (Xn /2) and the right boundary after this to (Xn) wouldn't we get a way better approximation ?(1 vote)
- It was just really rough approximation to introduce you to this topic. Of course you'd get much better approximation with middle or trapezoid boundaries.(3 votes)

- What is the mathematical difference between the Rectangular rule and Trapezoidal Rule?(1 vote)
- The video is clear about the physical difference. The mathematical difference is that the trapezoidal approximation usually results in an estimate of area that is closer to the actual value of the area.(2 votes)

## Video transcript

In the last few videos, we've
been approximating the area under the curve using
rectangles, where the height of each
rectangle was defined by the function evaluated
at the left boundary. So this would have been
the first rectangle. Then the second rectangle
would look something like this. And then we'd go all the
way to the nth rectangle would look something like that. And we saw-- so this
is the first rectangle, this is the second
rectangle, and we'd go all the way to
the nth rectangle-- and so we saw that the way that
you would take the sum of all of these rectangles in order
to approximate the area is that you would get the
sum from i equals 1 to n. And so i is essentially a
count of which rectangle we're dealing with. And what we're going
to do is multiply the height times the base. So the height of each rectangle,
the height of rectangle one, in this case, was the
function evaluated at x0. The height of rectangle two was
the function evaluated at x1. The height of rectangle
n was the function evaluated at x sub n minus 1. So the height of
a rectangle i is going to be the function
evaluated x sub i minus 1. If i is 2, then we're
evaluating it at x1. If i was 2, then this would be
the function evaluated at x1. So it's the left boundary. And then we have to
multiply it times the width. And in the last few
videos, and in this video, we will assume that all of the
rectangles have equal width. Now, we'll call that
equal width delta x. And to find it, we just have
to take the total distance that we're going
in the x direction. So it's going to be
b minus a divided by the number of
rectangles we want. So it's going to
be times delta x. Now, you might imagine that
this is not the only way to take the sum
using rectangles, or this is not the only
way to take the sum or approximate the area using
some type of geometric shape. For example, we could
have created rectangles where the height is defined
by the rightmost boundary. So let's define that. So here's our first rectangle. And we're defining the
height by the right boundary of the rectangle. So this right over
here is rectangle one, and its height is f of x1. And then for this
one right over here, we take the right boundary. The right boundary
defines that height. If we go all the
way to the-- this is rectangle two-- if we
go all way to the nth one, we use the right
boundary to define the height of the rectangle. So in this case-- this
is the nth rectangle-- how would we write this sum? Well, it would be
the sum-- which is, remember, we're just
trying to approximate the area under the curve-- from
i is equal to 1 to n. So i is a count of
each of the rectangles. And so the height of the
first rectangle is f of x1. The height of the nth
rectangle is f of x sub n. So this height right over
here is f of x sub n. So the height of
the i-th rectangle is going to be f of x sub i. Whatever the
rectangle number is, we take the x sub
that same number and evaluate the function there. That gives us the height. And we multiply
that times delta x. So the difference between
this and this here, for the i-th
rectangle, we use x sub i minus 1, so the left boundary. Here we use the right
boundary, f of x sub i. Well, we don't
have to stop there. Instead, we could use maybe
the midpoint between the two boundaries instead. So, for example,
over here we could we could use the midpoint
between x0 and x1 to find the height
of the rectangle. So this is right over here. This is f of x0
plus x1 over 2, just the midpoint between
these two points, to define the height
of the rectangle. So it would look
something like that. And the next one, we
would look at the midpoint to define the height. And we go all the
way to the nth one, and we define the
midpoint between its two sides of the rectangle. So the function
evaluated there tells us how high our
rectangle should be. And it would look
something like that. And so what would
this sum look like? Well, once again, we would
count each of our rectangles, so i equals 1 to
i equals n. i is which rectangle
we're working on. So this is the first one,
this is the second one, and this is the nth one. And the height isn't just
going to be f evaluated x sub i minus 1 or
f evaluated x sub i. It'll be the function
evaluated at the midpoint between the two-- x sub
i minus 1 plus x sub i, all of that over two,
and then times delta x. The delta x's are the same in
every one of these scenarios. Now finally, let's try to
break out of approximating only with rectangles and get a
little bit more creative. Why don't we try to
approximate with trapezoids? So let's try to do that. So what we could have here is
the left part of the trapezoid. The height is f of x sub 0. So this is f of x sub 0. And then the right side of
the trapezoid is f of x sub 1. And then what would be
the-- here, and let me do that for all of them. So that would be
the first trapezoid. Then the second trapezoid
would look like this. This one looks almost
like a rectangle, but we assume that the
top isn't completely flat. And then we go all the
way to the nth one. And this should be clear that
we're dealing with a trapezoid. All the way to the nth one
will look something like that. So how would we
calculate this area, the area of the trapezoid? Well, you just have to remember
that the area of a trapezoid is just the average of
the heights of the two sides times the base. So in this case-- and let me
write it out a little bit. So the area right
over there is going to be the average
of the heights. So it's going to be f of
x sub 0 plus f of x sub 1, all of that over 2. And then we're going to
multiply that times delta x. So that would be the area just
of this one right over here. We took the average
of the two heights and multiplied that
times the base. Now, if we wanted the sum of the
areas of all these trapezoids, and we wanted to write
in general terms, we could just write it's
the sum-- once again, we're going to count
the trapezoids. So this is the first
trapezoid, this is the second, all the way to
the nth trapezoid. So it's i equals
1 to i equals n. And the height of
each trapezoid, we're going to use the function
evaluated at the left boundary, x sub i minus 1, the average
of the function evaluated at the left boundary
and the function evaluated at the right
boundary, and we're going to take the
average of that and then multiply
that times the base. So the whole reason
why I wanted to do this is to show you there's
multiple, multiple ways of doing this. And in fact, if you wanted
to get really general, you could even have
different widths. But then that gets a
little bit more confusing. But really, just to
show you that you might see some of this fancy
notation in your calculus book or in your precalculus book. But all it's doing is summing
up the areas of trapezoids and rectangles depending
on whether they're using the right boundaries
of the rectangle to define the height, the
left boundaries, the midpoint of the left and the
right boundaries, or they could even
construct trapezoids.