### Video Transcript

In this video, we’ll learn how we
can estimate the area between a curve and the 𝑥-axis by splitting the region up
into rectangles. This is called a Riemann sum
approximation. We’ll discover how these
calculations can be greatly simplified using sigma notation and consider how more
complicated summation formulae might arise.

Let’s suppose we’re looking to find
the area between the curve of 𝑦 equals 𝑥 cubed, the 𝑥-axis, and the vertical
lines 𝑥 equals one and 𝑥 equals three. Now, there are ways that we can
evaluate the exact area. But it’s little outside the scope
of this lesson. Instead, we’re going to consider
how we can approximate the area between the curve and the 𝑥-axis. To do this, we have a number of
options. We’re going to use something called
Riemann sums or Riemann sum approximations. In this case, we split the area
under the curve into rectangles and find the area of each. And 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.

Okay, so let’s imagine we want to
split our area into four subintervals, four equally sized rectangles. The number of subintervals or
rectangles is denoted by the letter 𝑛. So here, we’re going to let 𝑛 be
equal to four. 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
divided by four, which is equal to 0.5. And each subinterval, which we call
𝛥𝑥, must be 0.5 units wide. Formally, we say that the width of
each rectangle 𝛥𝑥 is 𝑏 minus 𝑎 divided by 𝑛, where 𝑏 and 𝑎 are the endpoints
of our interval.

We’ll begin by using the right
endpoint of each subinterval for our estimation. In other words, we’ll find the
height of the rectangle by considering the function value at the right endpoint of
our subintervals. We saw that each rectangle is 0.5
units wide. So we add 0.5 to one. Then we see that the height of our
rectangle is equal to the value of the function when 𝑥 is equal to 1.5. Let’s call this 𝑓 of 𝑥 one. And that’s 𝑓 of 1.5. Now, our function is 𝑥 cubed. So 𝑓 of 1.5 is 1.5 cubed, which is
3.375. That’s the height of our
rectangle. The area of the rectangle is found
by multiplying its base by its height. That’s 𝛥𝑥 times the value of our
function 𝑓 of 𝑥 one. That’s 0.5 multiplied by 3.375
which is equal to 1.6875 square units.

Let’s repeat this process for our
next subinterval. We add another 0.5 to 1.5. This tells us the height of our
second rectangle is the value of the function when 𝑥 is equal to two. This time, we’ll say that’s 𝑓 of
𝑥 two, which is two cubed. Two cubed, of course, is eight. So the height of this rectangle is
eight units. This time, the area of the
rectangle is 𝛥𝑥 times this value of the function. That’s 0.5 times eight which, of
course, is four square units. Repeating this process once more,
we see that we need to evaluate the function when 𝑥 is equal to 2.5. That’s 2.5 cubed. And so the height of this third
rectangle is 15.625 units. Its area once again is 𝛥𝑥 times
this function value. That’s 0.5 multiplied by 15.625,
which gives us 7.8125 square units.

Adding another 0.5 and we get to
our fourth value of 𝑥; it’s three. That’s the 𝑛 of our interval and
the fourth rectangle as required. This time, 𝑓 of 𝑥 four or 𝑓 of
three is three cubed, which is 27. Its area will be given by its base
multiplied by its height. That’s 𝛥𝑥 multiplied by 𝑓 of 𝑥
four or 0.5 multiplied by 27 which is 13.5. The total area of these rectangles
and therefore an estimate to the area between the curve and the 𝑥-axis bounded by
the lines 𝑥 equals one and 𝑥 equals three is the sum of these four values. That gives us an area of 27 square
units. Now, 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
approximations more accurate by splitting the rectangles into smaller
subintervals.

It’s also important to remember
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 it. Although simply by evaluating the
function at these points, we’ll end up with negative values and, therefore, negative
values for the area. So that’s all fine and well. But is there a way we can formalize
this somewhat? Well, yes, there is. Let’s have a look of what we just
did.

Each time we multiplied the value
of 𝛥𝑥 by the value of the function at the sample endpoint. Here, it was the right endpoint,
but we could have chosen the left endpoint. And we’ll look at how that changes
our notation in a moment. We can write a general expression
for the approximation of the total area between the curve and the 𝑥-axis as 𝛥𝑥
times 𝑓 of 𝑥 one plus 𝛥𝑥 times 𝑓 of 𝑥 two all the way through to 𝛥𝑥 times 𝑓
of 𝑥 𝑛. We already saw here that 𝛥𝑥 is
equal to 𝑏 minus 𝑎 over 𝑛. We can also say 𝑥 one is 𝑎 plus
𝛥𝑥, 𝑥 two is 𝑎 plus two lots of 𝛥𝑥, all the way through to 𝑥 𝑛, which is 𝑎
plus 𝑛 lots of 𝛥𝑥.

But this is all still a little
messy. So we’re going to introduce a new
symbol. This symbol is sigma and it really
means “the sum of.” And it will help us neaten things
up a bit. We’re going to say that an estimate
for the area is the sum of all the 𝛥𝑥s times all the 𝑓 of 𝑥 𝑖s. Now, we saw that 𝑖 must start at
one and we saw that it ends at 𝑛. So we formalize this a little
further and we said 𝑥 𝑖 is equal to 𝑎 plus 𝑖 lots of 𝛥𝑥. So this is the estimate for the
area when we use a right Riemann sum. But what about when we use a left
Riemann sum? Well, when we write a right Riemann
sum, we take values of 𝑖 from one to 𝑛 and when we write a left Riemann sum we
take values of 𝑖 from zero to 𝑛 minus one. This essentially gives us the value
of our function at the left endpoint of each rectangle. We’re now going to look at a simple
application of these formulae before considering how it can help us to estimate the
area under a curve.

Represent the area under the curve
of the function 𝑓 of 𝑥 equals one over 𝑥 minus two in the closed interval 3 to 5
in sigma notation using a right Riemann sum with 𝑛 subintervals.

Remember, when we’re writing a
right Riemann sum, we take values of 𝑖 from one to 𝑛. The area is approximately equal to
the sum of 𝛥𝑥 times 𝑓 of 𝑥 𝑖 for values of 𝑖 between one and 𝑛. 𝛥𝑥 is 𝑏 minus 𝑎 divided by 𝑛,
where 𝑎 and 𝑏 are the lower and upper endpoints of our interval, respectively. And 𝑛, of course, is the number of
subintervals. Then 𝑥 𝑖 is 𝑎 plus 𝑖 lots of
𝛥𝑥. We always begin by working out
𝛥𝑥. We see that our closed interval is
from three to five. So we let 𝑎 be equal to three and
𝑏 be equal to five. 𝛥𝑥 is, therefore, five minus
three over 𝑛 which is, of course, two over 𝑛.

We’re now ready to work out what 𝑥
𝑖 is. It’s 𝑎, which we know to be three,
plus 𝛥𝑥, which is two over 𝑛, times 𝑖. Let’s just write that as three plus
two 𝑖 over 𝑛. Now, obviously for our summation,
we need to know 𝑓 of 𝑥 𝑖. 𝑓 of 𝑥 𝑖 must be 𝑓 of three
plus two 𝑖 over 𝑛. So lets substitute three plus two
𝑖 over 𝑛 into our formula. This gives us one over three plus
two 𝑖 over 𝑛 minus two which when we distribute our parentheses is simply one plus
two 𝑖 over 𝑛. This still isn’t particularly
nice. So what we’re going to do is
simplify the denominator.

We’ll write it as one over one plus
two 𝑖 over 𝑛. And then we’re going to multiply
the numerator and denominator of one over one by 𝑛. This creates a common denominator
of 𝑛 and then it means we can add the numerators. And we end up with 𝑛 plus two 𝑖
over 𝑛. Now over here, we’re working out
one divided by 𝑛 plus two 𝑖 over 𝑛. Well, another way to think about
that is think about its reciprocal. The reciprocal of 𝑛 plus two 𝑖
over 𝑛 is 𝑛 over 𝑛 plus two 𝑖. And we now have everything we need
to write our right Riemann sum. It’s a right Riemann sum, so we
start at 𝑖 equals one and we end at 𝑛. 𝛥𝑥 is two over 𝑛. And we multiply this by 𝑓 of 𝑥
𝑖, which we just worked out to be 𝑛 over 𝑛 plus two 𝑖. We then see that these 𝑛s
cancel. And we have our right Riemann sum
using sigma notation. It’s the sum of two over 𝑛 plus
two 𝑖 for values of 𝑖 between one and 𝑛.

In our next example, we’re going to
use the sigma notation to help us actually evaluate the area.

Compute the left Riemann sum for 𝑓
of 𝑥 equals one over 𝑥 squared plus two on the closed interval negative three to
three, given that there are six subintervals of equal width. Approximate your answer to two
decimal places.

Remember when we’re writing a left
Riemann sum, we take values of 𝑖 from zero to 𝑛 minus one. And that gives us the value of 𝑓
at the left endpoint of each rectangle. The formula is the sum of 𝛥𝑥
times 𝑓 of 𝑥 𝑖 for values of 𝑖 from zero to 𝑛 minus one where 𝛥𝑥 is 𝑏 minus
𝑎 over 𝑛. Remember 𝑎 and 𝑏 are the lower
and upper limits of our interval, respectively, and 𝑛 is the number of
subintervals. Then 𝑥 𝑖 is 𝑎 plus 𝑖 lots of
𝛥𝑥. It’s only sensible to begin by
first calculating 𝛥𝑥. We can see that our close interval
is from negative three to three. So we’re going to let 𝑎 be equal
to negative three and 𝑏 be equal to three.

We’re interested in six
subintervals. So let’s let 𝑛 be equal to
six. Then 𝛥𝑥 is three minus negative
three over six, which is simply one. Next, we’ll calculate what 𝑥 𝑖
is. It’s 𝑎, which we know to be
negative three, plus 𝛥𝑥, which is one, lots of 𝑖. That’s of course negative three
plus 𝑖. We’re looking to find what 𝑓 of 𝑥
𝑖 is though. 𝑓 of 𝑥 𝑖 must, therefore, be 𝑓
of negative three plus 𝑖. So let’s substitute 𝑥 equals
negative three plus 𝑖 into our function. That gives us one over negative
three plus 𝑖 squared plus two. And when we distribute the
parentheses, on the denominator we get 𝑖 squared minus six 𝑖 plus 11.

And now we’re ready to perform some
substitutions. We’re finding the sum and we’re
taking values of 𝑖 from zero to 𝑛 minus one. Now, 𝑛 is six. So 𝑛 minus one is five. 𝛥𝑥 is one and 𝑓 of 𝑥 𝑖 is one
over 𝑖 squared minus six 𝑖 plus 11. But of course, we don’t really need
to write multiplied by one. So we need to evaluate this
sum. To do this, we’re going to
substitute values of 𝑖 from zero through to five into this function and then find
their sum.

So when 𝑖 equals zero, that’s one
over zero squared minus zero plus 11. When 𝑖 is one, that’s one over one
squared minus six plus 11. When 𝑖 is two, it’s one over two
squared minus 12 plus 11. And we repeat this process for 𝑖
equals three, 𝑖 equals four, and 𝑖 equals five. The very last thing to do is to
evaluate their sum. That gives us 1.5909 and so on,
which are correct to two decimal places is 1.59. We’ve computed the left Riemann sum
for our function over that closed interval, using six subintervals.

In our very final example, we’re
going to look at how we deal with slightly more complicated summations.

Represent the area under the curve
of the function 𝑓 of 𝑥 equals 𝑥 squared minus one on the close interval zero to
three in sigma notation using a right Riemann sum with 𝑛 subintervals.

Remember when we’re finding a right
Riemann sum, we find the sum of 𝛥𝑥 times 𝑓 of 𝑥 𝑖 for values of 𝑖 from one to
𝑛. 𝛥𝑥 is 𝑏 minus 𝑎 over 𝑛 where
𝑎 and 𝑏 are the lower and upper limits of our interval, respectively, and 𝑛 is
the number of subintervals. 𝑥 𝑖 is 𝑎 plus 𝑖 lots of
𝛥𝑥. We always begin by working out what
𝛥𝑥 is. In our case, 𝑎 is equal to zero,
𝑏 is equal to three, and, well, 𝑛 is just 𝑛. This means 𝛥𝑥 is three minus zero
over 𝑛 or just three over 𝑛. Next, we’re going to work out what
𝑥 𝑖 is. It’s 𝑎, which we know to be zero,
plus 𝛥𝑥, which is three over 𝑛, times 𝑖. We’ll write this as three 𝑖 over
𝑛.

Of course, we want to know what 𝑓
of 𝑥 𝑖 is. So it follows that to find 𝑓 of 𝑥
𝑖, we find 𝑓 of three 𝑖 over 𝑛. Let’s substitute three 𝑖 over 𝑛
into our formula. That’s three 𝑖 over 𝑛 all squared
minus one which is nine 𝑖 squared over 𝑛 squared minus one. We’re now ready to use the
summation formula. We’re evaluating our sum for values
of 𝑖 from one to 𝑛. Its 𝛥𝑥, which is three over 𝑛,
multiplied by nine 𝑖 squared over 𝑛 squared minus one. We distribute our parentheses and
then we’re going to look to create a common denominator. We can do that by multiplying both
the numerator and denominator of our second fraction by 𝑛 squared. That gives us three 𝑛 squared over
𝑛 cubed, leaving us just to simply combine the numerators. We have 27 𝑖 squared minus three
𝑛 squared over 𝑛 cubed.

Now, we can simplify this
somewhat. The numerators share a factor of 27
and three. And of course they have a common
denominator of 𝑛 cubed. Both three and 𝑛 cubed are
indepnent of 𝑖. This means we can take three over
𝑛 cubed outside of the sigma symbol, and that means we’re done. We’ve represented the area under
the curve of the function in sigma notation with a right Riemann sum. It’s three over 𝑛 cubed times the
sum of nine 𝑖 squared minus 𝑛 squared for values of 𝑖 from one 𝑛.

In this video, we’ve learned that
we can estimate the area between the curve and the 𝑥-axis by splitting the region
into rectangles. We saw that we can use the sigma
symbol to mean this sum of, and that provides us with the right Riemann sum on the
left Riemann sum formulae. Remember, when we’re dealing with
the right Riemann sum, we take values of 𝑖 from one to 𝑛. And when we’re dealing with the
left Riemann sum, we take values of 𝑖 from zero to 𝑛 minus one.