INTEGRATION & RIEMANN SUMS
For our purposes, the goal of this subject is to find the area under a graph between two x-values, a and b:
We do this by finding rectangles whose areas add up to being close to the area under the graph, and then making the rectangles narrower and narrower so that more and more of them fit under the curve between a and b. We find that the narrower and more numerous we make them, the closer their total area comes to equaling the actual area under the graph.
Let’s start with two rectangles, and agree that their bases will divide the interval [a,b] exactly in half, and agree to choose their heights so that their right edges just touch the curve.
Every time we divide the area up into rectangles, we’ll call the number of rectangles N and the lengths of the rectangle bases “dx.” So, dx will always = (b-a)/N and every time we increase the number of rectangles (N goes up), dx will also change (go down).
In this case, N = 2 and dx = (b-a)/2, that is, (b-a)/N.
Also, it makes the formulas easier later if we put xi labels on the x values where the rectangles touch the x-axis, as shown. x0 always = a, and xN always = b, (and N is still the number of rectangles.).
What is the area of the 1st rectangle? height*width = f(x1)*dx
What is the area of the 2nd rectangle? height*width = f(x2)*dx
The total area of those two rectangles, the gray area, is sort of close to the area we want, but pretty far off. It is:
A = f(x1)*dx + f(x2)*dx
It will help later, if we also notice now what the “xi” actually are:
x0 = a
x1 = a+1*dx
x2 = a+2*dx = b
The rest of the process is easier to understand if we agree that no matter how finely we divide up the x-axis (making narrower and more numerous rectangles between a and b), we will continue to use the same method to decide where they go and their dimensions. That is, we will always divide the x-axis into equal-length segments, and always choose the rectangle heights so that their right edges just touch the curve.
So, now we’ll use 4 rectangles and see what happens.
In this case, N = 4 and dx = (b-a)/4, that is, (b-a)/N. Again, note that a and b are constants but N and dx are variables and changed when we increased the number of rectangles.
Look how much closer the gray area is to what we want than it was with just two rectangles. The new area of rectangles is:
A = f(x1)*dx + f(x2)*dx + f(x3)*dx + f(x4)*dx
Note again here what the “xi” actually are:
x0 = a
x1 = a+1*dx
x2 = a+2*dx
x3 = a+3*dx
x4 = a+4*dx = b
At this point, notice two things:
The more rectangles we use, the closer is their total area to that under the graph. That means that we are probably going to want to use a large number of them.
If we do use a lot of them, the formula for “A” will get very long and it will get very hard to write it, read it, copy it, or do much of anything with it.
To solve this problem with the formula for “A,” mathematicians use a shorthand called “summation notation.” It is handy when you have a long sum with terms that are very similar except for possibly one number that is counting up in each one. Here is an example of summation notation:
Now, notice:
The left hand side means, “substitute every i from 3 to 8, inclusive, into the expression xi, then add them all up.
The left-hand side of the equal sign means the same thing as the right-hand side, but is a lot easier to handle.
Here are a few more examples of summation notation, including some we will need:
Now, notice:
The index does not have to be “i”
The second example (with “3pk”) shows that you can move any factors (like 3 and p) that don’t involve the index out in front of the summation.
The third example (ending in “4n”) shows that even if the index does not appear in the expression at all, you still add the expression up once for each time the index counts from its starting to its final value.
Now, we can use this summation notation to write our formulas for area. Here is the one we did for N=4 in “long notation” and summation notation:
A = f(x1)*dx + f(x2)*dx + f(x3)*dx + f(x4)*dx
Now, look how easy it is to change the formula to represent the area of any number of rectangles. Here is the area if we divide it up into 1000 rectangles:
x0 = a
x1 = a+1*dx
x2 = a+2*dx
:::
x1000 = a+1000*dx = b
As the number of rectangles gets bigger and bigger, the sum A does not get bigger and bigger but begins to home in on a specific value, which is, in fact, the actual area under the graph. In these formulas, you cannot see that happening, but it does. So, each of the following 4 example sums is getting closer and closer to that actual area (note that the only differences in the formulas as you move down the list are that N is increasing and dx is decreasing):
That is, using ten thousand rectangles gets closer, but one hundred thousand rectangles will get still closer, a million rectangles closer still, and ten million rectangles even closer, and so forth.
Looking at all the formulas above for total rectangle area, I hope you can see that the one for “N” rectangles, where N is any integer, is:
x0 = a
x1 = a+1*dx
x2 = a+2*dx
:::
xN = a+N*dx = b
Putting all this together, we see that we can make the sum as close as we like to the actual area under the graph by just making N large enough. We express this fact as a limit like this:
In the last formula, since dx does not contain or depend on the index “i,” we can pull it out in front of the summation sign:
Finally, though it has been more compact and convenient to call the rectangle width “dx,” it will be easier to calculate limits of actual functions if we now write for dx for what it actually is, in terms of N, like this:
Where:
xi = a+i*dx
Note:
Now we can clearly see that “dx” changes when N changes
(b-a) can be pulled in front of the limit sign because it does not contain “N”, the variable under the limit sign.
Finally, let’s apply this last result to a specific problem, the area under the function
f(x)=1 + 3x between -1 and 5. That is:
(Equation 1)
because:
f(xi) = 1 + 3xi
We can start by evaluating just the summation, like this:
(Equation 2)
To get any further, we need a formula for xi. The formula will use the limits of integration specific to the integral we are evaluating (a = -1 and b = 5).
We earlier said that: xi = a+i*dx
We also know that dx = (5-(-1))/N = 6/N, so,
xi = -1+i*(6/N)
So, that last summation in Equation 2 becomes:
If we put this back in Equation 2, we get:
If we put this back in Equation 1, we get:
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a
b
AREA IN HERE
b
a
b
a
x0
x1
x2
x0
x1
x2
x3
x4