The limit we spent 5.1 (the limit of as n approaches infinity of the sum of a function times the change in x for n changes) that popped up in the area under a curve and the distance on object travels apparently pops up in a bunch of other applications later down the road. This name is the Definite Integral.
Definite Integral - For a function f, if it is defined between x values a and b, we can divide it into n sections of equal length [(b-a)/n]. If xv0=a and xvn =b, and xv1...xvn are any sample points within the segment that lie between the actual segments (xv(i-1)) then the definite integral of f from a to b is | ⌠(a to b)f(x)(dx) = Lim<n→∞> Σ(i=1→n) f(xi*)▲x | If this limit exists, then f is integrable on [a,b].
The limit is then plugged into the "precise definition of a limit" which I don't feel like replicating here.
As a side note, the book points out that the elongated s was chosen as the integral symbol because it is a limit of sums. f(x) is the integrand, and a and b are the limits of integraion. Calculating an integral is called integration, not to be confused with the civil rights movement, though it makes civil rights speeches funny to assume that's what is being discussed.
Side note 2: the definite integral is a number and the symbols are arbitrary. Whether f(x)dx uses an x, a t, or a picture of Pikachu, the value of the integral does not change.
Side note 3: The sum of the right-hand side of the equation is called a Riemann Sum after the German guy who figured it out.
So, we can say that the definite integral of a function is the limit of its Riemann Sum and can be calculated to however closer of an answer you want.
It's a little more flexible than the sum of the rectangles thing through, because, even though it looks like the exact same thing if they're all positive, it changes a bit if f has negative values. When f dips under the x axis, then the Riemann sum is the areas of the positive parts of the curve minus the area of the negative parts, or, as the book puts it, the net area. So the Riemann sum is Av1 - Av2, assuming A2 is negative and A1 is positive (and delicious).
The book adds another side note that sometimes we don't use equal subintervals , and that you can also define the limit as when ▲xi → 0 instead of as n → ∞. I guess we'll work on that later.
Just to reassure us, the book points out a theorem that, as long as f is continuous or has a finite number of jump discontinuities on its interval, the function is integrable. So, as long as it never stretches to inifnity and leaves a gap in your interval, you're good.
Since the easiest place to take your sample points is usually the right endpoint of the segments and being able to do it from any point in the middle is mainly a parlor trick or a way to break up monotony if for some reason you have to integrate all day, we can redefine the definite integral so that the Rieman Sum is from i=1 to n for f(xi)▲x if we're using segments of equal length.
The example problem demonstrates that, most of the time, when rewriting a Riemann sum as an integral, we just replace the Limit and Sigma symbols with the Integral symbol, the x with the star above it with just x, and the ▲x with dx.
Since we're going to be given Rieman sums or Integrals, then asked to evaluate them, we lay out some ground rules to work with them that will save you hours of tinkering with the limit-taking process. I'm not typing them out mathematically, since I don't have the ALT codes memorized yet and it wouldn't do me much good. Instead, I'm going to try to summarize them in words, since I just had to go look up Sigma notation to figure out exactly how to understand what the heck I've been reading. So, here are the formula (formulas? Formulae? Formuleaux?) for dealing with sum.
1. The sum of n number of integers is half the product of n and (n + 1).
2. The sum of the squares of n number of integers is 1/6th the product of n, n+1, and 2n+1.
3. The sum of the cubes of n number of integers is the square of the sum of the integers formula.
4. The sum of n number of the the same constant is n times the constant.
5. The sum of n numbers of a constant times a function's value is the same as the constant times the sum of the functions. (The n can be factored out by the distributive property)
6. The sum of two series of numbers is the same as the sums of each of them individually, added together.
7. The sum of the difference of two series of numbers is the same as the difference of each of their sums. (That doesn't make a lot of sense written out. If you heard me say it out loud and point at it, it would totally make sense.)
A couple of examples demonstrate how to evaluate limits, and we move on to a section called The Midpoint Rule.
Most of the time, the right endpoint of the ith segment of an interval is the easiest choice for i to compute the interval. But sometimes we want to find an approximation (Instead of an exact value which could be found with the right endpoint. I'm sure the reasoning for this will be apparently later, but it seems weird now.) The symbol for our sample point is an i with a star above it, but the midpoint is denote by x, with a subscript of i, and and line over it, becuase they want this to be as annoying as possible to type about.
When using the midpoint, we use the definite integral, but ▲x is is .5(xv<i-1>+xvi), or the point that is in the middle of each segment. Through the example problem, they show that using that method on an earlier example gets a much closer estimate for the chosen n than using the right endpoint.
So, if you don't want to bother evaluating the integral, using the midpoint of the segments is a better estimate than either of the endpoints. Two chapters from now we'll apparently learn how to estimate the error and learn some new ways to approximate an integral.
The next section develops some basic properties, which I will once again summarize instead of trying to type.
1. The integral of a constant function is the constant times the length of the segment.
2. The integral of the sum of two function is the sum of the integrals of the two functions
3. The integral of a constant times a function is equal to the constant times the integral of the function. (The constant can be factored out)
4. The integral of the difference of two functions is equal to the difference of their integrals.
5. The sum of the integrals of two adjacent segments equals the integral of the entire segment.
I'm not sure why the book decided to stick number 3 between 2 and 4. Whatever. Pretty straightforward.
Also, my book has a typo, I believe. For rule five, they have listed that the integral from a to c plus the integral from c to b equals the integral from a to b, which doesn't make any sense. My teacher walked through it and wrote it down in the way that made sense, so I'll double-check this with her, but I think the book is wrong here.
Oh, just after writing that, I looked more closely at their diagram, and they just decided to label the segments weird. With the way they labeled them, their formula works. Why they they made c the point in between a and b is a mystery to me. Maybe I'm just getting tired and overthinking this.
Those properties were true for any comparison of a and b (though I don't see how the adjacent sections property. There a few more properties that only work if a ≤ b and x is between them or equal to them.
6. If f(x) ≥0 then the integral ≥ 0, or, if the function is positive, the integral is positive.
7. If f(x) ≥ g(x), then the integral of f(xdx) ≥ the integral of g(x)dx. Big functions have big integrals. Little functions drive big cars to compensate.
8. if m ≤ f(x) ≤ M , then the the integral lies between m times the segment length and M times the segment length.
Property 8 is referring to the absolute minimum and absolute maximum, which the book decides to explain after listing it all and pretending this is perfectly standard notation and wondering why you'd ever be confused by this.
Anyway, Prop 8 (ha) says that the integral is going to be bigger than the area of the rectangle taken with the height of the absolute minimum, and smaller than the area of the rectangle made with the height of the absolute maximum. Not entirely sure why this is useful knowledge, but I'm sure it will be obvious later and I will regret doubting the wisdom of their theorems, since we've all pretty much been doing that since we wondered why we would EVER need to know how to do fractions back in Grade School.
Well, according to the book, it can be used for a quick estimate if you don't feel like using the midpoint rule. Okay, sure.
Oi. Finally done with this section. It's late and I'm tired, but I'm gonna try to do at least some of the homework before class on Monday and I know I won't get any done tomorrow. I may try a little tonight, or I may just wake up a little earlier tomorrow and try to do stuff before church and socializing takes over the day. But I probably won't, because I know that when I wake up early tomorrow homework will be the last thing I want to do. I guess I'll dive in, and probably try to finish anything I don't get done on Monday before class, since my 7 AM class won't be meeting that day.
Okay, good night.
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