This section is still under construction
Return with me to the days of yore when, if the local miller didn't grind enough grain into flour, the whole community would go hungry. In the village I am picturing, there is but one mill. A dam holds back a brook to form a millpond. The water from the pond turns a waterwheel, which turns the mill, which grinds the grain. The demand for flour is so great that the miller and his sons work shifts to keep the mill productive twentyfour hours a day. So what limits production is how fast the pond water can turn the waterwheel. And that rate is in direct proportion to how high the water level is in the pond.
The miller needs to project his inventory far into the future in order to serve his customers well. To do this, he needs to know how much production he can get in any period. That period might be any week, any month, any quarter, or an entire year.
The problem is, the pond's water level varies seasonally. Based upon past weather patterns, the miller has determined that the water level that the water level, in meters, is
x(t) = 1.5 + cos( (pt/182.5)  1.2) eq. 10.31where t is days past January first. Until now the miller's method of calculating production for any period has been to plot the water level as a function of time throughout that period on finely ruled graph paper. Then he simply counts the number of squares under the graph. He multiplies that by a conversion constant and out comes his production.
But counting all those tiny little squares on the graph paper got to be a strain on the old man's eyes, not to mention his patience. One evening his eldest son, who was learning calculus in his time off from the mill, saw his dad sweating over next quarter's projections.
"What you doin', Dad?" he asked. "Drat!" the miller exclaimed. "You've made me lose count." "Count of what?" So the miller explained the procedure to his son. "Why don't you just use antiderivatives?" the son asked. It was clear from the expression on the old man's face that it was the son's turn to explain things now. "Ok Dad. The fundamental theorem of calculus tells us that if you pick an antiderivative of your x(t) function it will be some area function of that same x(t). It might not be the exact area function you are interested in, but it's easy to make it into the one you're interested in." 

Pink area is proportional to production for the January through March quarter.  
Yellow area is proportional to production for the April through June quarter. 
"And how do you do that?" asked the dad.
"Look here. I can tell, because I know my derivatives, that an antiderivative
of your
X(t) = 1.5t + (182.5/p)sin((pt/182.5)  1.2) + C eq. 10.32where C is some undetermined constant. You can see that if you take the derivative of this X(t) you get your x(t). "
"Well that's just great," the dad complained. "I'll go and tell the village elders and the mortgage company that production numbers for next quarter are subject to an undetermined constant."
But Dad, it doesn't matter what C is. An antiderivative gives you the area under your curve from some fixed time past until a time of your choosing. Let's say that we chose C so that it gave it to us from the first of the year. You are trying to figure numbers for the second quarter. Now watch. I'll color the first quarter area on your graph pink and the second quarter area yellow. If I take X(t) for t at the end of the first quarter, it will give me the pink area. If I take it for t at the end of the second quarter it will give me the pink area plus the yellow area. So to get the yellow area, I merely subtract the first number from the second. So the first quarter ends at the end of the 90th day of the year. The second quarter ends at the end of the 181st day of the year. So the yellow area must be
A_{yellow} = X(181)  X(90)It's that easy."
"So what happened to the undetermined constant," asked the dad.
"Well just substitute in the expression for X(t) and you'll see.
A_{yellow} = (1.5 × 181) + (182.5/p)sin((181p/182.5)  1.2) + C _{ }  (1.5 × 90)  (182.5/p)sin((90p/182.5)  1.2)  CThe C cancels out exactly. So you don't have to care what it is. It will always cancel."
"But what if the pink area doesn't start at the beginning of the year? What about that, Mr. TooMuchEducation?"
"It still doesn't matter. You're just asking what would happen if the pink area started at a different time. But you will always be taking the pink plus the yellow and subtracting away the pink. The pink always cancels no matter where it starts. The difference will always give you the yellow area. Whatever your yellow area is, just take an antiderivative at the t where it ends and subtract that same antiderivative at the t where it begins. All you need to do is plug your choice of antiderivative function into your pocket calculator and you've got the production projections."
"Just one problem," said the dad. "We're living in the days of yore. Pocket calculators won't be invented for another 300 years."
The function, X(t), shown in equation 10.32 is called the indefinite integral of the x(t) shown in equation 10.31. The notation the world uses to represent this is
1.5 + cos((pt/182.5)  1.2) dt = 

1.5t + (182.5/p)sin((pt/182.5)  1.2) + C eq. 10.33 
But if we do as the miller's son suggested and evaluate the indefinite integral
at both
181 90 
1.5 + cos((pt/182.5)  1.2) dt = 171.2889333 meterdays eq. 10.34 
Observe the units of the answer. The units of the integrand function is meters, because it is measuring how high the water in the millpond is. The units of the span over which we took the definite integral is days, which is also the units of the dt operator. The units of the definite integral will always be the product of the units of the integrand with the units of the independent variable. So here we get meters times days. You will notice that that is also the units of area on our graph (figure 10.31)  that is the horizontal axis is in days, the vertical axis is in meters, so any rectangle in that graph has area with units of meters times days.
You can always distinguish between a definite integral and an indefinite integral because only the definite integral has limits above and below the integral sign.
The procedure for evaluating a definite integral is
Here is an example: evaluate
3 2 
x^{2} dx_{ } 
^{ } x^{3} x^{2} dx = 
A convenient notation for doing all this in one line is:
3 2 
^{ } x^{2} dx = _{ } 
x^{3} 
3 ^{ } 2 
27 8 19 = 
Suppose instead of using the antiderivative to find the area under the
curve, f(x), between
(b  a) Dx =If you want to know the x for the lefthand edge of the kth rectangle, it would beeq. 10.35 n
x_{k} = a + (k1)Dx eq. 10.36The area of the kth rectangle would be
A_{k} = f(x_{k})Dx = f(a + (k1)Dx)Dx eq. 10.37So the full Riemann sum, done out for you is
^{ } n A_{total} = å f(a + (k1)Dx)Dx eq. 10.38a _{ } k=1The green expression in the above is, according to equation 10.36, just another way of expressing x_{k}. So 10.38a is really just
^{ } n A_{total} = å f(x_{k})Dx eq. 10.38b _{ } k=1Now look at the beige expressions. When
^{ } bDx A_{total} = å f(x_{k})Dx eq. 10.38c _{ } x_{k}=aNow we take the limit as n goes to infinity. I will show that by replacing the blue thing by a different blue thing.
^{ } A_{total} = _{ } 
bDx x_{k}=a 
f(x_{k})Dx eq. 10.38d 
The other Dx is the red one. We are multiplying f(x_{k}) by it. As n goes to infinity we are adding up more and more products of f(x_{k})'s with Dx. The Dx in turn gets smaller and smaller. So I am going to change the name of Dx to dx to indicate that it is not just a very small quantity, but rather an infinitessimal quantity  that is a quantity so close to zero (closer than any nonzero real number) that we can only tell it apart from zero by adding up infinitely many of them. And that's what we do, but first we multiply each dx by a selection from the infinite collection of f(x_{k})'s that we have. And by the way, the k in the x_{k} is a counting variable. Since you can't count to infinity, in the limit let's just drop it and use x by itself. So now we have
^{ } A_{total} = _{ } 
b x=a 
f(x)dx eq. 10.38e 
Another name for the infinitessimal quantity, dx, is a differential quantity or more often just a differential. This concept of an infinitessimal quantity is one that is often difficult for first year students to become comfortable with. So here's another way of looking at it, this time from the antiderivative point of view.
If F(x) is an antiderivative of f(x), then
dFBut what does that mean? Remember our definition of a derivative? Here it is again, but this time using Dx in place of the h that we used back in section 4.= f(x) eq. 10.39a dx
F(x + Dx)  F(x) limwhich is just saying that f(x) is the derivative of F(x). Now instead of taking the limit right away, let's suppose only that Dx is very close to zero. This means that the two sides of equation 10.39b will only approximate each other now:= f(x) eq. 10.39b Dx>0 Dx
F(x + Dx)  F(x)Now multiply both sides by Dx:» f(x) eq. 10.39c Dx
F(x + Dx)  F(x) » f(x) Dx eq. 10.39dNotice that the righthand side of this is just the area of the rectangle whose height is f(x) and whose width is Dx.
Now suppose that the xaxis is broken up into segments, each of width Dx. And lets give each segment an index, k. So the left endpoint of the kth segment is x_{k}. The same approximation still holds for x_{k}:
F(x_{k} + Dx)  F(x_{k}) » f(x_{k}) Dx eq. 10.39e
But isn't
F(x_{k+1})  F(x_{k}) » f(x_{k}) Dx eq. 10.39f
So what happens if we add up these expressions for a bunch of consecutive segments? Let's say we do it from a segment whose left endpoint is a to another whose left endpoint is b.
b b å F(x_{k+1})  F(x_{k}) » å f(x_{k}) Dx eq. 10.39g x_{k}=a x_{k}=a
Look first at what happens on the left. In every case the F(x_{k+1}) from one summand will cancel with the F(x_{k}) of the summand for the next k  in every case, that is, except the first summand and the last summand. From the first summand the uncanceled term will be the F(x_{k}). But that x_{k} is the left endpoint of the first segment, and we called that a. So the contribution we get from it is F(a). The term from the other end that remains uncanceled will be the F(x_{k+1}) from the last summand. We know that the x_{k} for this summand is the left endpoint of the last segment, which is at b, so what we really have here is F(b+Dx). So putting in what we know about the sum on the left of equation 10.39g we have:
b F(b+Dx)  F(a) » å f(x_{k}) Dx eq. 10.39h x_{k}=aThe righthand side is just another Riemann sum. So when we take the limit as the number of summands goes to infinity and Dx goes to zero (becoming the infinitessimal quantity, dx), the F(b+Dx) on the left goes to F(b), and the sum on the righthand side goes to the definite integral (and again we drop the k because it can't count to infinity):
F(b)  F(a) = 
b x=a 
f(x) dx eq. 10.310 
Here is a shortened version of the odessy we just took. We started with
dFWe multiplied through by the infinitessimal quantity, dx (that was in equation 10.39d).= f(x) dx
dF = f(x) dxNotice we have infinitessimal quantities on both sides now. Then we summed up an infinitude of them over the interval, a to b. When you sum up an infinite number of infinitessimals, that's an integral. We found that on the left everything cancelled except
Using the antiderivatives for the various functions we have already discussed in the previous section (see the table to find the antiderivatives), evaluate the following definite integrals:
a) 
121 4 
x^{1/2} dx 
b) 
3 0.5 
2^{x} dx 
c) 
2 5 
x^{2} dx 
d) 
5 5 
3x + 4 dx 
e) 
b 0 
mx dx 
Section 11.1: Intro to Methods of Integration still under construction