### Video Transcript

In this video, we’ll look to define
the definite integral of a function formally as the limit of a Riemann sum. In doing so, we’ll establish how we
can express definite integrals as limits of Riemann sums and vice versa. And we’ll evaluate a definite
integral by taking the limit of the corresponding Riemann sum written in sigma
notation.

Remember, we can estimate the area
between the curve and the 𝑥-axis, which is bounded by the lines 𝑥 equals 𝑎 and 𝑥
equals 𝑏, by splitting the region into, say, 𝑛 rectangles and calculating the area
of each. This is called finding a Riemann
sum. And it’s defined using sigma
notation as the area is approximately equal to the sum of 𝑓 of 𝑥𝑖 star times Δ𝑥
for values of 𝑖 from one to 𝑛. Here Δ𝑥 is equal to 𝑏 minus 𝑎
over 𝑛. Contextually, that gives us the
width of each of our rectangles. And 𝑥𝑖 star is any sample point
in the subinterval 𝑥𝑖 minus one to 𝑥𝑖.

Now, of course, it follows that as
𝑛 increases, the width of each of our rectangles gets smaller. And this results in a more accurate
estimate for the area. In fact, as 𝑛 approaches ∞
— the number of rectangles approaches ∞ — the limit of this sum approaches
the exact area of the region.

We can therefore say that the area
required is equal to the limit as 𝑛 approaches ∞ of the sum of 𝑓 of 𝑥𝑖
star times Δ𝑥 for values of 𝑖 from one to 𝑛. And in fact, this limit is hugely
important as it occurs in a wide variety of situations, even when 𝑓 is not a
positive function. We therefore give it a special
name, 𝑛 notation.

We come to the definition of a
definite integral. We say that if 𝑓 is a function
defined for 𝑥 is greater than or equal to 𝑎 and less than or equal to 𝑏, we can
divide the closed interval 𝑎 to 𝑏 into 𝑛 subintervals of equal width. We let 𝑥𝑖 star be the sample
points in each subinterval such that 𝑥𝑖 star lies in the closed interval 𝑥𝑖
minus one to 𝑥𝑖. Then we say that the definite
integral of 𝑓 from 𝑎 to 𝑏 is the limit as 𝑛 approaches ∞ of the sum of 𝑓
of 𝑥𝑖 star times Δ𝑥, values of 𝑖 from one to 𝑛. And that’s, of course, provided
that this limit exists and gives the same value for all sample points. If it does exist, we say that 𝑓 is
integrable on the closed interval 𝑎 to 𝑏.

Now this integration symbol here
was introduced by Leibniz. It’s an elongated S and was chosen
because the integral is the limit of the sums. Now note that not all functions are
integrable, although most commonly occurring functions are. In fact, if 𝑓 is continuous on the
closed interval 𝑎 to 𝑏 or if it has only a finite number of jump discontinuities,
then 𝑓 is integrable on the closed interval 𝑎 to 𝑏. In other words, the definite
integral from 𝑎 to 𝑏 of 𝑓 of 𝑥 with respect to 𝑥 exists.

Now in fact if 𝑓 is integrable on
the closed interval 𝑎 to 𝑏, then this limit must exist. And it must give the same answer no
matter which value, which sample point 𝑥𝑖 star, we choose. We can therefore simplify our
calculation by choosing purely right 𝑛 points. We can say that if 𝑓 is integrable
on the closed interval 𝑎 to 𝑏, then the definite integral between 𝑎 and 𝑏 of 𝑓
of 𝑥 with respect to 𝑥 is the limit as 𝑛 approaches ∞ of the sum of 𝑓 of
𝑥𝑖 times Δ𝑥 for values of 𝑖 from one to 𝑛. And here Δ𝑥 is 𝑏 minus 𝑎 over
𝑛, and 𝑥𝑖 is 𝑎 plus 𝑖 lots of Δ𝑥.

This is a definition we’re going to
be using throughout the remainder of this video. And we now have everything we need
to be able to express definite integrals as limits of Riemann sums and vice
versa.

Express the definite integral
between three and nine of three 𝑥 to the sixth power with respect to 𝑥 as the
limit of Riemann sums.

Remember, if 𝑓 is integrable on
the closed interval 𝑎 to 𝑏, then the definite integral between 𝑎 and 𝑏 of 𝑓 of
𝑥 with respect to 𝑥 can be expressed as the limit of Riemann sums as shown. Let’s compare everything in our
theorem to our integral. Our function is a polynomial. And we know that all polynomials
are continuous over their domain, which means that they are therefore integrable
over their domain. So the function 𝑓 of 𝑥 equals
three 𝑥 to the sixth power is continuous and therefore integrable over the closed
interval defined by the lower limit three and the upper limit nine.

So we’re going to let 𝑎 be equal
to three and 𝑏 be equal to nine. We’ll move on and define Δ𝑥. It’s 𝑏 minus 𝑎 over 𝑛. Well, we said 𝑏 is nine and 𝑎 is
three. And this is all over 𝑛. That gives us that Δ𝑥 is equal to
six over 𝑛. And we can now define 𝑥𝑖. It’s 𝑎 plus 𝑖 lots of Δ𝑥. Well, 𝑎 is three. And we need 𝑖 lots of Δ𝑥, which
we worked out to be six over 𝑛. Let’s write that as three plus six
𝑖 over 𝑛.

In our limit, we’re going to need
to work out 𝑓 of 𝑥𝑖. So it follows that we can find this
by substituting the expression for 𝑥𝑖 into our function. That gives us three times three
plus six 𝑖 over 𝑛 to the sixth power. And we can now substitute
everything we have into our definition for the integral. When we do, we see that the
definite integral between six and nine of three 𝑥 to the sixth power with respect
to 𝑥 is equal to the limit as 𝑛 approaches ∞ of the sum of three times
three plus six 𝑖 over 𝑛 to the sixth power times six over 𝑛 evaluated between 𝑖
equals one and 𝑛.

Since multiplication is
commutative, we can rewrite three times six over 𝑛 as 18 over 𝑛. And we have our definite integral
expressed as the limit of Riemann sums. It’s the limit as 𝑛 approaches
∞ of the sum of 18 over 𝑛 times three plus six 𝑖 over 𝑛 to the sixth power
evaluated from 𝑖 equals one to 𝑛.

Let’s now have a look at a more
complicated example.

Without evaluating the limit,
express the definite integral between negative five and two of the square root of
seven minus four 𝑥 squared with respect to 𝑥 as a limit of Riemann sums.

Remember, if 𝑓 is integrable on
some closed interval 𝑎 to 𝑏, then the definite integral between 𝑎 and 𝑏 of 𝑓 of
𝑥 with respect to 𝑥 is equal to the limit as 𝑛 approaches ∞ of the sum of
𝑓 of 𝑥𝑖 times Δ𝑥 for values of 𝑖 from one to 𝑛. We calculate Δ𝑥 by subtracting 𝑎
from 𝑏 and then dividing by 𝑛. And 𝑥𝑖 is 𝑎 plus 𝑖 lots of
Δ𝑥.

In this case, we can say that 𝑓 of
𝑥 is equal to the square root of seven minus four 𝑥 squared. The lower limit of our integral is
negative five. So we’ll let 𝑎 be equal to
negative five and the upper limit is two. So let’s let 𝑏 be equal to
two. It’s always sensible to next work
out Δ𝑥. Here, that’s 𝑏 minus 𝑎. So that’s two minus negative five
all over 𝑛. That gives us Δ𝑥 to be equal to
seven over 𝑛. Then we can work out 𝑥𝑖. It’s 𝑎, which is negative five,
plus 𝑖 lots of Δ𝑥, which we just calculated to be seven over 𝑛. This simplifies to negative five
plus seven 𝑖 over 𝑛.

Next, we work out 𝑓 of 𝑥𝑖. And it follows that we can find
this by substituting 𝑥𝑖 into the expression for 𝑓 of 𝑥. It’s the square root of seven minus
four times negative five plus seven 𝑖 over 𝑛 squared. And we now have everything we need
to be able to express our limit. We replace Δ𝑥 with seven over 𝑛
and 𝑓 of 𝑥𝑖 with the square root of seven minus four times negative five plus
seven 𝑖 over 𝑛 all squared.

And so we obtain that our integral
as a limit of Riemann sums is the limit as 𝑛 approaches ∞ of the sum of
seven over 𝑛 times the square root of seven minus four times negative five plus
seven 𝑖 over 𝑛 squared for values of 𝑖 from one to 𝑛. Now this example is interesting as
this equation isn’t actually integrable over the given interval. Remember, for 𝑓 to be integrable,
it must be continuous over the interval 𝑎 to 𝑏. Well, the graph of 𝑦 equals the
square root of seven minus four 𝑥 squared looks a little something like this. This is clearly not continuous over
the closed interval negative five to two. So we wouldn’t be able to evaluate
this limit.

It’s important to realise that we
can carry out the procedure and get a Riemann sum. But we do need to check that the
function is integrable over that region. Although we have a resulting
expression in this case, it isn’t a valid answer to the question. If we’d instead been given limits
of, say, negative square root of seven over two and positive square root of seven
over two, then it would’ve been fine. Let’s now see how we can reverse
the process and express a limit in integral form.

Express the limit as 𝑛 approaches
∞ of the sum of 𝑒 to the power of 𝑥𝑖 over two minus four 𝑥𝑖 times Δ𝑥𝑖
for values of 𝑖 from one to 𝑛 as a definite integral on the closed interval
negative five to negative three.

Remember, if 𝑓 is integrable on
some closed interval 𝑎 to 𝑏, then the definite integral between 𝑏 and 𝑎 of 𝑓 of
𝑥 with respect to 𝑥 is equal to the limit as 𝑛 approaches ∞ of the sum of
𝑓 of 𝑥𝑖 times Δ𝑥 for values of 𝑖 from one to 𝑛. Now we can quite clearly see that
our interval is from negative five to negative three inclusive. So we begin by letting 𝑎 be equal
to negative five and 𝑏 be equal to negative three.

Let’s now compare our limit to the
general form. We can see that 𝑓 of 𝑥𝑖 is equal
to 𝑒 to the power of 𝑥𝑖 over two minus four 𝑥𝑖. Well, that’s great because that
means 𝑓 of 𝑥 is equal to 𝑒 to the power of 𝑥 over two minus four 𝑥. This means the limit of our Riemann
sums can be expressed as a definite integral. It’s the definite integral between
negative five and negative three of 𝑒 to the power of 𝑥 over two minus four 𝑥
with respect to 𝑥.

In our final example, we’ll look at
how we can actually evaluate the integral by calculating the limit of Riemann
sums.

Evaluate the definite integral
between negative four and two of negative 𝑥 minus four with respect to 𝑥 using the
limit of Riemann sums.

Remember, if 𝑓 is some integrable
function on the closed interval 𝑎 to 𝑏, then the definite integral between 𝑎 and
𝑏 of 𝑓 of 𝑥 with respect to 𝑥 is defined as the limit as 𝑛 approaches ∞
of the sum of 𝑓 of 𝑥𝑖 times Δ𝑥 for values of 𝑖 from one to 𝑛. And of course, Δ𝑥 is 𝑏 minus 𝑎
over 𝑛 and 𝑥𝑖 is 𝑎 plus 𝑖 lots of Δ𝑥.

Let’s begin by comparing this
definition with our integral. We see that 𝑓 of 𝑥 is equal to
negative 𝑥 minus four. This is a polynomial. And We know polynomials are
continuous over their domain. So the function negative 𝑥 minus
four is continuous and therefore integrable over the closed interval defined by the
lower limit negative four and the upper limit two.

We therefore let 𝑎 be equal to
negative four and 𝑏 be equal to two. And we’ll next look to define
Δ𝑥. It’s 𝑏 minus 𝑎. So that’s two minus negative four
over 𝑛. That gives us six over 𝑛. Next, we’ll define 𝑥𝑖. It’s 𝑎, which is negative four,
plus 𝑖 lots of Δ𝑥, which we calculated to be six over 𝑛. This gives us that 𝑥𝑖 is negative
four plus six 𝑖 over 𝑛.

It follows that we can find 𝑓 of
𝑥𝑖 by substituting this expression into our function. That gives us negative negative
four plus six 𝑖 over 𝑛 minus four. When we distribute the parentheses,
we find 𝑓 of 𝑥𝑖 to be equal to four minus six 𝑖 over 𝑛 minus four. And of course, four minus four is
zero. So 𝑓 of 𝑥𝑖 is simply negative
six 𝑖 over 𝑛.

And we can now substitute
everything we have into our definition for the integral. This tells us that the definite
integral between negative four and two of negative 𝑥 minus four with respect to 𝑥
is equal to the limit as 𝑛 approaches ∞ of the sum of negative six 𝑖 over
𝑛 times six over 𝑛 for values of 𝑖 from one to 𝑛. Now in fact, negative six 𝑖 over
𝑛 times six over 𝑛 is negative 36𝑖 over 𝑛 squared. So this is the sum.

Now of course, this factor,
negative 36 over 𝑛 squared, is actually independent of 𝑖. So we can rewrite our limit as the
limit as 𝑛 approaches ∞ of negative 36 over 𝑛 squared times the sum of 𝑖
from values of 𝑖 from one to 𝑛. And in fact, whilst it’s outside
the scope of this video to prove this, we can quote the general result. The sum of 𝑖 from 𝑖 equals one to
𝑛 is equal to 𝑛 times 𝑛 plus one over two. And this means we can replace all
of this sum with the expression 𝑛 times 𝑛 plus one over two.

Our definite integral can therefore
be evaluated by working out the limit as 𝑛 approaches ∞ of negative 36 over
𝑛 squared times 𝑛 times 𝑛 plus one over two. And you might spot that we can
divide both 36 and two by two. We can also cancel one 𝑛. And our limit reduces to the limit
as 𝑛 approaches ∞ of negative 18 times 𝑛 plus one over 𝑛.

Let’s distribute our
parentheses. And when we do, we find that this
can be written as negative 18𝑛 over 𝑛 minus 18 over 𝑛. Now of course, negative 18𝑛 over
𝑛 is just negative 18. And we can now evaluate our limit
by direct substitution. As 𝑛 approaches ∞, negative
18 over 𝑛 approaches zero. And so the limit as 𝑛 approaches
∞ of negative 18 minus 18 over 𝑛 is simply negative 18. We can therefore say that the
definite integral between negative four and two of negative 𝑥 minus four with
respect to 𝑥 is negative 18.

In this video, we’ve seen that we
can write a definite integral as the limit of Riemann sums. We say that, for an integrable
function 𝑓 over some closed interval 𝑎 to 𝑏, the definite integral between 𝑎 and
𝑏 of 𝑓 of 𝑥 with respect to 𝑥 is equal to the limit as 𝑛 approaches ∞ of
the sum of 𝑓 of 𝑥𝑖 times Δ𝑥 for values of 𝑖 from one to 𝑛. Here Δ𝑥 is 𝑏 minus 𝑎 over
𝑛. And 𝑥𝑖 equals 𝑎 plus 𝑖 lots of
Δ𝑥.

We saw that we can use this
definition to write an integration as a limit of a summation and vice versa. And that, with some clever
manipulation, we can even evaluate these limits to help us calculate the
integral.