- Infinite series
The jet engine fan blades on the left (below) and the single windmill blade on the right are three dimensional compound curves, designed for maximum efficiency and the immense strength needed to do their jobs. These complicated shapes have to be modeled mathematically, input into simulators, tweaked when new information or new materials become available, and they have to be formed using computer controlled robots because they're just too difficult for a free-hand machinist to make and reproduce.
Integral calculus and particularly the definite integral is the mathematics of such curves. It allows us to determine cross-sectional areas, volumes and surface areas of odd shapes, and much much more. Here we go ...
Photos: Rolls Royce, ReCharge.com
We're going to go at the definite integral first from a geometric perspective. That part is pretty easy; the integral represents the area between a curve and the x-axis (we'll have to think about that more carefully later), and between domain points a and b, that is, in the interval [a, b]. We'll calculate that area by dividing it up into small rectangles, and in the fashion of calculus, increasing our accuracy by decreasing the size and adding rectangles, even "to infinity" to get at the exact area. We'll look at the geometric interpretation for some time before shifting to the analytic one for most of our future work.
In the analytic representation, we just have to introduce the limits of integration, a and b, attached to the integral sign, below.
We'll develop the two-part Fundamental Theorem of Calculus in this section. It will allow us to take a huge shortcut in the calculation of the areas under and between complicated curves.
While it might be tempting (and probably correct) to suggest that the true area under the curve for a given number of rectangles might be something like an average of these three, we don't really need to do that. It turns out that any of these approaches will work, so long as we let the width of the rectangles shrink to zero and their number grow to infinity.
When we add these areas, the sums will be called the left-, right- and middle Riemann Sums, respectively, after the 19th century mathematician who developed this way of thinking.
Using the Left-Riemann picture, we set the width of each rectangle to Δx, where Δx = (b-a)/N, and N is the number of rectangles. The heights of the rectangles are f(a), f(a + Δx), f(a + 2Δx), and so on. The sum is:
where L stands for "left.
Similarly, we can define the Right- and Middle-Riemann sums:
Now we formally suggest that as n → ∞, all of these sums approach the area class the curve so divided into rectangles:
We need to work through one example of Riemann-sum integration of a "complicated" function in order to see how it works. Doing two more carefully chosen examples will help to make the transition between Riemann integrals and doing definite integrals using the fundamental theorem of calculus. The examples are:
The first of these will take a while, so have a beverage and get ready. The second two are very easy and take about a minute each to understand. The pattern that emerges will tell us a valuable story. We're letting the lower limit be zero for convenience, but we can show later that it can be any number. Let's go.
To find the area under f(x) = x2 between x=0 and x=1, we divide into rectangles, as usual. We'll employ a right-Riemann sum here (because it's convenient and it doesn't matter which flavor we use).
The width of each rectangle is b/N, which will approach zero as N→ ∞. The height of the first three rectangles are shown. The sum of these areas for a fixed N is:
Now, in a couple of steps, we'll plug in what we know from the diagram (remembering that f(x) = x2):
Recognizing that (b/N)3 is in every term, we can simplify this to:
Now what to do about the infinite sum, 12 + 22 + 32 + . . . + N2 ? First, it's not going to be infinite if it's divided by N3 (which is hidden in that (b/N)3 term), so there's some hope that our formula for the area A will be bounded (won't be infinite). In order to figure this out, we can take a well known digression ... to Egypt, if you will.
Consider a pyramid built of 1 x 1 blocks. A top and side view are shown below. The volume of the base, a square layer of 1 x 1 blocks, is N blocks on a side, with volume N2. The second layer up has volume (N-1)2, the third (N-2)2, and so on.
The volume of a general square pyramid, of the pyramid formed by tracing the lower corners (green lines) of each layer and the upper corners (red lines) of each layer are shown. Notice that the actual volume of the pyramid (which is the sum 12 + 22 + 32 + . . . + N2, must be less than the volume of the red pyramid and greater than that of the green one.
That notion gives us the inequality
Dividing by N3 and making a slight rearrangement of the right side gives:
Now if we take the limit as N→ ∞, the limit on the right is just 1/3:
Well, all of that pyramid stuff led us right to what we needed. The value of that sum divided by N3 is just 1/3.
Now we've shown that
It took a while, but we won't have to do any (many?) more examples of full-on Riemann integration like this, but first ... our other two examples.
This definite integral is considerably simpler because we know that the area of a triangle (this is a right triangle) is
Because f(b) = b, we have
... and that's it. That's the integral. Now for the third and final example.
The area under this simple curve is just 1 x b = b.
Also notice that if we made the left hand limit of this integration a number, a, instead of zero, that area would be A = b - a.
Because of our convenient choice of zero for our starting point (a) we've overlooked an important part of the pattern, but we'll recover that below. The main result is that doing a Riemann sum is the same as finding an indefinite integral, then evaluating it at the limits of integration, a & b.
This is a remarkable result, and worth repeating. The value of a definite integral, which represents the area under a curve between two points in the function domain, is calculated by performing a Riemann sum. However, that process can be cut short by finding the indefinite integral of the function, F(x) (dropping the constant of integration) and evaluating the result between the two limits of integration: F(b) - F(a)
It's worth introducing some new notation at this point.
In this example, f(x) = x2 is integrated again, but this time between the limits of 3 and 5. Make sure you can follow the simple procedure (which is a vast shortcut to Riemann sums).
Now compare that last integral with the definite integral of f(x) = x3 between x=3 and x=5. The procedure is the same, just find the antiderivative of x3, F(x), then evaluate between the limits by subtracting F(3) from F(5).
Also notice in this example that x3 > x2 for all positive x, and the value of the integral is larger, too.
Here's a slightly more complicated example, but as you can see, it is solved in the same way.
All of the properties of the indefinite integral, including that the integral of a sum is the sum of integrals of the parts, and that a multiplicative constant can be removed from the integral, still hold (see the table below).
The definite integral, however, has limits, and those bring in new properties. An important one is illustrated on the right. If c lies in [a, b], then the integral of a function between a and b can be broken into the sum of integrals from a to c and from c to b.
The negation property (2) is interesting. If F'(x) = f(x) in the example in the table, then we are comparing F(b) - F(a) to F(a) - F(b). Makes sense.
This begs the question: What do we mean by "area under the curve"? Between 0 and π, this integral has a value of 1, the area under that part of the curve. If we calculate the integral between π and 2π using the fundamental theorem of calculus, however, we get -1, and the sum is zero.
Not to worry , there are two different ways to think about this and they're both right, depending on the question you're asking. First, let's focus on area between the curve and the x-axis. This suggests that "negative" areas simply need to be subtracted from positive areas (or alternatively, we used absolute values), like this:
The other way to think of this integral is as a cumulative sum. Say the sine function corresponds to velocity as a function of time. Then its integral is the sum of infinitesimal distances traveled between time = 0 and time = 2π. In that case, the net distance traveled is zero. After all the same path was taken in the negative direction as in the positive, and whatever was moving (like maybe the end of a spring) has ended up right back where it started after one sine cycle.
In integrating f(x) = x3 - 4x2 -4x + 16 between -2 and 4, with the explicit goal of determining the shaded area, we first have to know the location of the zero at x=2. From there it's a straightforward matter to break the into two pieces, the first the integral of the positive area between -2↔2, and the negative area 2↔4.
Here's the calculation. Don't lose sight of the fact that calculating odd areas like this amounts to little more than finding an easy antiderivative and doing some arithmetic.
Calculate each of the following integrals as a cumulative sum. That means just evaluate the indefinite integral between the two endpoints. For some (your choice), you should calculate the absolute value of the area between the curve and the x-axis. For example, for problem #1 you would integrate from -2 to 0 and from 0 to 3 and add the absolute values of the two integrals.
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