In the most recent tutorial there is a question about converting a Riemann sum to a definite integral, and it seems to be tripping up quite a few students. I wanted to run through one of the calculations in detail so you can see how to answer such a question.

Let’s look at the example:

There are many ways to tackle such a question but let’s take one particular path. Let’s start by the fact that when the limit is defined, the limit of a sum is the sum of the limits. We can split up our expression into 3, which looks like:

Let’s tackle each of these separately. Let’s look at the first term:

Well, we can take the factor of 13 outside the front of the whole thing to start with, along with the factor of 9, and this will give

We see here that we have a sum of terms, and a factor which looks like in each term. We notice however that inside the bracket there is a , so we try to make the term on the outside look the same as this. Multiply and divide by 6 to give:

Now, the is a constant (as we sum over the i’s), and will correspond to the width of the rectangles in the area which is ultimately what a Riemann sum tells us about. Remember in a Riemann sum, we want a function value at a particular point, times a width. We will call the width of each rectangle . So we can write:

Now, in a Riemann sum, we have a function evaluated at various points, which we generally call . As we increase i by 1, we move to the next rectangle in our approximation. Here we see a factor of (pulling out the factor of 6). Remembering that is the width of our rectangles, when i=1, this is going to give the point 0, and when i=n, this is going to give the point . That’s not very interesting, but if we were to go to the next point (i=n+1), we would have which would be the point 6. So, running through the values from i=1 to i=n, the quantity goes from the value 0 to the value just below 6, where as we take n to infinity, we get closer and closer to 6 for that last point. We can treat as our , so long as we know that we are going from to . So, we can write:

with to

This can be converted directly into the definite integral:

Exactly the same procedure works with the other expressions in the sum, and we end up with:

where the 117 comes from 13 x 9, the 104 comes from 13 x 8 and the 91 comes from 13 x 7.

The last integral we can write as:

(Make sure that you can see that this is true).

And so we can actually write the whole thing as a single integral:

Which can be calculated from the known antiderivatives and the fundamental theorem of calculus part 2.

prof dr mircea orasanloiosuNovember 15, 2017 at 4:13 pmindeed is very impoortant the questions

liettew claireFebruary 16, 2018 at 5:15 ami wish you actually showed how you changed the terminals of integration rather than just telling us we should see that it’s true.

Jonathan ShockFebruary 16, 2018 at 6:09 am@Liettew Claire, have you tried it? Give it a go. I leave it like that because it is something that you can explore, and find for yourself. If it still doesn’t make sense, I am happy to lead you through it.

liettew claireFebruary 16, 2018 at 8:53 pmi had to draw boxes to understand this but i dont get how it works generally for non constant integrands

Jonathan ShockFebruary 16, 2018 at 9:03 pmLet’s take the integral of the function f(x)=x from x=0 to x=6. If you wanted to change it to an integral from 0 to 1 you would want to define a new variable, related to x, which, when x was 6, was equal to 1 and when x was 0 was equal to zero. So, this new variable could be, for instance p=x/6. Then x=6p. We can change variables now to p. the function we are integrating over becomes simply 6p, and dx becomes 6 dp. So now we have the integral over (6p 6 dp) from p=0 to p=1. This is 36 times the integral over p dp from p=0 to p=1. Which is 36 times p^2/2 from p=0 to p=1, which is 36 times 1^2/2 which is 36. Now check that indeed this is the same as the original x integral.

Does that make sense? Try it for the integral over x^2dx.

liettew claireFebruary 16, 2018 at 9:14 pmyes it makes sense. not sure how this principle works for constant integrands though, how does it work if f(x) = 2 and i want F(x) over the interval [0,3]?

can i even do a substitution in this case ?

Jonathan ShockFebruary 16, 2018 at 9:17 pmAbsolutely…Give it a go and see…check that the final answer is the same, whatever substitution you make, so long as you make the substitution in the variable, and in the limits

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