Video Transcript
In this video, we’re going to learn
how definite integrals can be estimated by using rectangles. And this is a process called
finding Riemann sums. We’ll discover how to split the
area between the curve and the 𝑥-axis into rectangles of varying points. And consider how accurate these
approximations are by looking at a number of examples of various degrees of
difficulty.
Let’s suppose we’re looking to find
the area between the curve of 𝑦 equals 𝑥 squared, the 𝑥-axis, and the vertical
lines 𝑥 equals one and 𝑥 equals three. We know that we can find the exact
area using definite integration. We evaluate the integral of 𝑥
squared between the limits of one and three with respect to 𝑥. The integral of 𝑥 squared is 𝑥
cubed over three. Evaluating this between the limits
of one and three, and we obtained three cubed over three minus one cubed over three,
which is 26 over three or roughly 8.67 square units. The problem is, integrating a
function is not always that simple. And we might just need to
approximate the area between the curve and the 𝑥-axis.
To do that, we have a number of
options. And in this video, we’re going to
use something called Riemann sums. In this case, we’re going to split
the area under the curve into rectangles and then find the area of each. There are three ways we could do
this. We could find the height of the
rectangles by using the value of the function at the left endpoint of each
rectangle, at the right endpoint of each rectangle, or at the midpoint of each
rectangle. And on occasion, we might even
choose to use unequally sized subintervals, though that’s quite rare. Let’s go back to this example.
We’re going to imagine we want to
split the area into four subintervals, four equally sized rectangles. To find the width of each
rectangle, we find the difference between the bounds of our area. And we share that into 𝑛 pieces,
where 𝑛 is the number of subintervals. Here, that’s three minus one all
divided by four, which is of course one-half. Each sub interval is therefore 0.5
units wide. Formaly, we say that the width of
each rectangle, 𝛥𝑥, is equal to 𝑏 minus 𝑎 over 𝑛. Where 𝑏 and 𝑎 are the endpoints
of the interval and 𝑛 is the number of subintervals. Let’s estimate the area by using
the right endpoint of each sub interval. In other words, we’re going to find
the height of each rectangle by considering the function value at the right endpoint
of each of our subintervals.
Starting at the beginning of our
interval, we add 0.5 to one. And we see that the height of this
rectangle is equal to the function value at 𝑥 equals 1.5. That’s 𝑓 of 1.5, which is of
course 1.5 squared, which is equal to 2.25. Since the area of a rectangle is
found by multiplying the length of its base by its height, we find the area of this
one by multiplying 0.5 by 2.25. Which gives us 1.125 square
units. Next, we add another 0.5 to 1.5 to
give us two. The height of this rectangle is the
value of the function at 𝑥 equals two. That’s 𝑓 of two which is two
squared which is, of course, equal to four. This time, the area of the
rectangle is four times 0.5, which is two square units. We add then another 0.5 to two to
give us 2.5. And when we see that the height of
this rectangle would be the value of 𝑓 at 2.5. 𝑓 of 2.5 is 2.5 squared, which is
6.25. And so, the area of our third
rectangle is 6.25 times 0.5, which is 3.125 square units.
Finally, if we add another 0.5, we
get to three. That’s the ends of our
interval. And it’s the fourth rectangle as
required. This time, its height is 𝑓 of
three. That’s three squared, which is
nine. And therefore, the area of our
final rectangle is nine times 0.5, which is 4.5 square units. To find an estimate for the total
area under the curve and therefore an estimate to the definite integral between one
and three of 𝑥 squared, we add these together. An estimate then to that integral
is 1.125 plus two plus 3.125 plus 4.5, which is 10.75 square units.
Remember, the answer that we got
when we evaluated the integral itself was 8.67. And if we look carefully, we can
see that our rectangles are all a little bit bigger than the area required. So we would have been expecting an
overestimate. We could, of course, make our
approximation more accurate by splitting the rectangles into smaller
subintervals. And, in fact, as the number of
subintervals 𝑛 approaches infinity, the Riemann sum approaches the definite
integral of the function between the endpoints. It’s important to remember, though,
that if the function takes on both positive and negative values as shown here. Then the Riemann sum is the sum of
the areas of the rectangles that lie above the 𝑥-axis and the negatives of the
areas of the rectangles that lie below the 𝑥-axis. In this case, that would be the
area of the pink rectangles minus the area of the yellow rectangles. We’re now going to look at a more
formal example of this process, this time using left endpoints.
Given that 𝑓 of 𝑥 is equal to
four cos 𝑥 and that 𝑥 is greater than or equal to zero and less than or equal to
𝜋 by four, evaluate, to the nearest six decimal places, the Riemann sum for 𝑓 with
six subintervals, taking the sample points to be left endpoints.
Remember, we can approximate the
definite integral of a function between the limits of, say, 𝑎 and 𝑏 by splitting
the area between the curve and the 𝑥-axis into 𝑛 subintervals. Where the width of each rectangle,
𝛥𝑥, is equal to the difference between the upper and lower limits divided by 𝑛,
the number of subintervals. In this question, we’re looking to
split the area into six subintervals, so we let 𝑛 be equal to six. We let 𝑎 be equal to zero and 𝑏
be equal to 𝜋 by four. So 𝛥𝑥 here and the width of each
of our rectangles is 𝜋 by four minus zero over six, which is 𝜋 over 24. Let’s now sketch out the curve of
𝑦 equals four cos 𝑥 between 𝑥 equals zero and 𝜋 by four radians and spilt it
into these six subintervals.
This question requires us to use
left endpoints. So the height of each of our
rectangles will be equal to the value of the function at the left of each
subinterval. The sketch is really useful here
because we see that each rectangle sits above the 𝑥-axis. So we just need to find the
positive area of each. Let’s consider our first
rectangle. Its left endpoint is at 𝑥 equals
zero. So we find the height of the
rectangle by evaluating 𝑓 of zero. 𝑓 of zero is four times cos of
zero, which is four. The area of this first rectangle
then is therefore 𝜋 by 24, since that’s the width of each rectangle, times
four. That’s 𝜋 by six square units.
We now need to find the next
endpoint. We achieve this by adding 𝜋 by 24
to zero; that’s 𝜋 by 24. And we can therefore find the
height of this rectangle by evaluating 𝑓 of 𝜋 by 24. That’s four cos of 𝜋 by 24, which
is approximately 3.965779 and so on. The area of our second rectangle
then is width times height or base times height. That’s 𝜋 by 24 times 3.965 and so
on, which is roughly 0.51911931 and so on. Now, of course, this question is
asking us to evaluate till six decimal places, which is why I’ve left the numbers as
I have. We find the next endpoint by adding
another 𝜋 by 24 to 𝜋 by 24 and we get 𝜋 by 12. The height of this third rectangle
then is 𝑓 of 𝜋 by 12 which is four cos of 𝜋 by 12, which gives us root six plus
root two. And then we see that the area of
our third rectangle is 𝜋 by 24 multiplied by this value, which gives us an area of
roughly 0.505 and so on square units.
We repeat this process by
repeatedly adding 𝜋 by 24 to each of our endpoints. And when we do, we find that the
left endpoint of each of our rectangles is 𝜋 by eight, 𝜋 by six, and 𝑥 equals
five 𝜋 by 24. Their heights are respectively
given by 𝑓 of 𝜋 by eight, 𝑓 of 𝜋 by six, and 𝑓 of five 𝜋 by 24. And when we multiply each of these
values by the width, 𝜋 by 24, we find that the area of our fourth rectangle is
roughly 0.4837. Our fifth rectangle is roughly
0.4534. And our sixth is approximately
0.4153. Remember, the Riemann sum is the
total area of these six rectangles. It’s 2.901 and so on. That’s 2.901067 correct to six
decimal places.
In our next example, we’re going to
look at how to find the Riemann sum using midpoints.
Given 𝑓 of 𝑥 is equal to 𝑥
squared minus four and 𝑥 is greater than or equal to negative four and less than or
equal to two, evaluate the Riemann sum for 𝑓 with six subintervals, taking sample
points to be midpoints.
Remember, we can approximate the
definite integral of a function between the limits 𝑎 and 𝑏 by splitting the area
between the curve and the 𝑥-axis into 𝑛 subintervals. The width of each rectangle is
given by 𝛥𝑥, which is found by subtracting 𝑎 from 𝑏 and dividing this value by
𝑛. In this example, we want to split
the area into six subintervals. So we let 𝑛 be equal to six, and
then 𝑎 is negative four and 𝑏 is two. The width of each of our
subintervals is given then by two minus negative four over six, which is equal to
one. Next, we’re going to sketch the
curve of 𝑦 equals 𝑥 squared minus four between the endpoints negative four and
two. And we’re going to split this into
six subintervals.
This question requires us to use
midpoints. So the height of each rectangle
will be equal to the value of the function at the middle of each subinterval. That’s going to look a little
something like this. Now, we can work out the 𝑥-value
at the end of each rectangle by repeatedly adding one to four. And that gives us each of these
values. We can then see that the midpoints
are going to be negative 3.5, 𝑥 equals negative 2.5, negative 1.5, and so on.
Now, we’re going to need to be
extra careful here. We see that some of our rectangles
sit below the 𝑥-axis. This means we’re going to find the
negative value of their area. In other words, we’ll subtract the
total area of the rectangles that sit below the 𝑥-axis from the total area of the
rectangles that sit above the 𝑥-axis. Let’s begin by finding the function
values at each midpoint. And that will tell us the height of
each rectangle. That’s 𝑓 of negative 3.5, 𝑓 of
negative 2.5, 𝑓 of negative 1.5, and so on. And, of course, our function is 𝑥
squared minus four. And when we substitute each of
these values into that function, we get 8.25, 2.25, negative 1.75, negative 3.75,
another negative 3.75, and another negative 1.75. We then obtain the area of our
first rectangle here to be 8.25 times one.
Our second rectangle has an area of
2.25 times one. Our third rectangle has an area of
1.75 times one, not negative 1.75. Because we’re just dealing with
areas at the moment. Our fourth rectangle has an area of
3.75 times one. Base times height with our fifth
rectangle is 3.75 times one again. And for our last rectangle, it’s
1.75 times one. The Riemann sum is therefore 8.25
plus 2.25 minus the sum of 1.75, 3.75, 3.75, and another 1.75. And that gives us an approximation
to the definite integral between the values of negative four and two of 𝑥 squared
minus four. It’s negative 0.5. And, of course, 𝑥 squared minus
four is a fairly simple function to integrate. So we can check our answer by
evaluating that integral.
When we do, we get 𝑥 cubed over
three minus four 𝑥 between the limits of negative four and two, which gives us a
value of zero. And our estimate of negative 0.5 is
pretty close. So we can assume we’ve probably
done this correctly. It’s worth noting that a sketch of
the curve won’t always be possible. So instead, we need to notice that
when the value of 𝑓 of 𝑥 is less than zero, we subtract the area of the rectangle
with that height.
Now that we’ve considered three
types of Riemann sums, let’s have a look at how to decide whether we’re finding a
lower or upper estimate for our integral.
A table of values for an increasing
function 𝑓 is shown. Use Riemann sums to find lower and
upper estimates for the definite integral between the limits of zero and 20 of 𝑓 of
𝑥 with respect to 𝑥.
It’s important to know that we’ve
been given a table of values for an increasing function. This means that over the closed
interval zero to 20, the graph of the function will always be sloping up. It might look a little something
like this. Now, we know we can find estimates
for the integral given by using Riemann sums. So our job here is to work out
which Riemann sum is likely to give the lowest estimate and which is likely to give
the highest estimate. If we were to find the Riemann sum
using the left endpoints, we see we end up with something a little like this. Almost every rectangle is a little
smaller than the actual area. When we use right endpoints, we end
up with much more than the actual area. It follows also that using the
midpoint would give us something in the middle as would using the trapezoidal
rule.
This means that using the left
endpoint is going to give us a lower estimate and using the right endpoint will give
us an upper estimate. Let’s find both of these
approximations. We see that the width of our
subintervals is four. So we could say that the width of
each rectangle is going to be four units. And the area of our first rectangle
is going to be four multiplied by the positive value of the function at the left
endpoint; that’s three. The second rectangle will have an
area of four times four. The third will have an area of four
times nine. Then we have a rectangle with an
area of four times 18 and an area of four times 30.
Remembering that the first
rectangle sits below the 𝑥-axis, we subtract this from the sum of the other
four. And we get a total area of 232
square units. We now repeat this using the right
endpoint. And we see that the area of our
first rectangle is four times four. The second is four times nine. Then, we have an area of four times
18, four times 30, and four times 35. This time, each of these rectangles
sits above the 𝑥-axis. So we add all these values together
to get a total of 384 square units. The lower and upper estimates of
the definite integral of 𝑓 of 𝑥 evaluated between zero and 20 are therefore 232
and 384, respectively. In fact, we can generalize the
result here. We can say that when working with
increasing functions. The left Riemann sum will be an
underestimate of the area between the curve and the 𝑥-axis. Whereas the right Riemann sum will
give an overestimate. The reverse, of course, must be
true then for decreasing functions.
Suppose 𝑇, 𝐿, and 𝑅 are
approximations for the definite integral between one and three of 𝑥 cubed plus one
using the trapezoidal rule, left Riemann sum, and right Riemann sum, respectively,
using 10 subintervals. Which of the following is the
correct relationship between the three approximations? Is it 𝑅 is less 𝑇 which is less
than 𝐿, 𝐿 is less than 𝑇 which is less than 𝑅, 𝑅 is less than 𝐿 which is less
than 𝑇, or 𝑇 is less than 𝐿, which is less than 𝑅?
Let’s consider the function 𝑓 of
𝑥 equals 𝑥 cubed plus one. We know by considering the shape of
the curve that it’s an increasing function. Though we could check this by
differentiating to get three 𝑥 squared which is greater than zero for values of 𝑥
not equal to zero, confirming our suspicions. This really helps because we know
that for an increasing function, the left Riemann sum gives an underestimate for the
actual area. Let’s call that 𝐴. And the right Riemann sum gives an
overestimate for the actual area. We can therefore say that we know
that 𝐴 must be greater than 𝐿 which must be less than 𝑅. And this, in turn, means that 𝑅,
of course, must be greater than 𝐿.
But where did the estimates for the
trapezoidal rule come in here? Well, this uses nothing more than
the average of the left and right Riemann sums. So it follows that this
approximation must sit between the two. And we obtain that 𝑇 must be
greater than 𝐿 but less than 𝑅. The second is the correct
relationship between the three approximations.
In this video, we learned that we
can approximate definite integrals by splitting the area between the curve and the
𝑥-axis into a number of rectangles. Each rectangle have heights of the
value of the function at the subinterval’s right endpoints, left endpoints, or
midpoints. We also saw that for increasing
functions, the left Riemann sum will give a lesser value than the right Riemann
sum. And for decreasing functions, the
reverse is true.
