Video: Definite Integrals as Limits of Riemann Sums

In this video, we will learn how to interpret a definite integral as the limit of a Riemann sum when the size of the partitions tends to zero.

14:49

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.

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