Video: Riemann Sums and Sigma Notation

In this video, we will learn how to use sigma notation with Riemann sums to find the area under a curve.

14:25

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.

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