We got our review quiz back, and I had a 77 on it. Looking back over it, it was mainly careless errors and forgetting my trig identities, so it's annoying but not devastating. Anyway, on to 5.4
5.3 took a lot of mental effort to process, but, from what I've read so far, 5.4 is just some applications of it, so it shouldn't be quite as taxing. That's a good thing, because I have a quiz to do after the homework for this section. Fun times.
To make it easier to write down, we're using ⌠f(x)dx as the notation for an antiderivative of f(x). Since the integral on a segment is the definite antiderivative, using the integral without a given segment is an indefinite integral, and will be a general antiderivative with a +C. If you have a given segment, the definite integral gives a number - an absolute area. If you use the indefinite integral, you get a function.
We then have a table of indefinite integral formulas, which are pretty much the opposite of the derivative formulas + C, so I'm not going to bother summarizing them.
Also, the indefinite integral is only valid on an interval. It compares it to the general antiderivative, but I don't fully understand the difference it's showing. Something to ask about later.
Now, since ⌠f(x)dx is the antiderivative of the function, then ⌠F'(x) is the derivative of the antiderivative of function, so it's just another way of saying that function. But, looking at it as the derivative of the antiderivative, it becomes the rate that the antiderivative changes as x changes.
So stating our our ⌠f(x)dx = F(b)-F(a) as ⌠F'(x)dx = F(b)-F(a) makes its use in science applications more apparent, since we know that the rate the integral changes is the same as the net change between the interval.
Example: if C(x) is the cost of producing x units of something, then we know the marginal cost (C'(x)) is the rate of change in the cost. So ⌠C'(x)dx = C(x2)-C(x1) would give us the increase in cost from one amount to the other.
There are a few more examples, but that's pretty much all their is to this chapter. A littler easier to wrap my head around it.
Tuesday, January 25, 2011
Sunday, January 23, 2011
5.3 - The Fundamental Theorem of Calculus
Two semesters into your math degree and we're now going to teach you the foundation of what you've been studying. That explains all the creaking and shaking.
The two kinds of calculus we've been doing - integral differential, seem unrelated at first glance, but, as Newton and his crack team of mathematicians discovered, they're closely related. The two processes are actually the inverse of each other, and that will enable us to not do these long processes from the last two chapters except on tests from now on. At least, that's what I'm hoping.
We start this off by defining a function as g(x) = ⌠(a to x) f(t)dt, where f is a continuous function on [a,b] and x can be either a or b. The value of g is only dependent on x, which is the top number of our interval. If x is fixed, then the integral is a constant, but if x is a variable, then the integral varies and is a function of g(x).
If f(t) is positive, then g(x) is the area under the graph of f from a to x, and x can vary. g(x) is the "area so far," as the book puts it, which I like.
This is the first entry where I'm doing the writing before having the class on the topic, so I don't fully understand what's going on, but I'll write my best understanding of it and probably talk about my clarifications after class.
Anyway, we're shown a curvy graph for y=f(t) and, given that the integral from 0 to x of that function is a function of x, evaluate that function [g(x)] for a series of numbers. We can see from looking at it that g(0) will be 0, and the rest of them can be interpreted as areas of fairly easy geometric shapes. Except a couple of them, which they eyeball to say are about 1.3, because I guess that's okay here.
We plot these points on a graph of g(x). Since the areas switched to negative areas at one point, the graph begins decreasing there and we know we have a maximum. Using an exercise my teacher didn't assign so I had to look up, we know that the integral of a variable function is half of the difference between b^2 and a^2, which, if we choose a as 0 and b as x, we get (x^2/2) for the integral, and thus g(x).
Now, the book says that we take f(t) to equal t, which I don't fully understand the reasoning behind and will ask about in class tomorrow. But, just taking their word from, if we take the derivative of g(x), we get x. So, we can say g'=f, meaning that g is an antiderivative of f.
Anyway, if we estimate the derivative of g(x) and sketch our graph out, it should look pretty close to the graph we started with, which confirms our suspicion.
To make it more mathematically sound than "Hey, those graphs look like each other," we think about any graph where f(x)≥0, so that g(x) (being the integral from a to x of f(t)) is the area under the graph from a to x. If we're going to compute g'(x) using the limit-taking method none of us remember how to do because we haven't used it since section 2, using g(x) and g(x+h) actually results in us subtracting areas. Since we're subtracting the area from the area of h further than our first area, what we really get is the area of the rectangle whose width is h, and whose height is f(x) for that point, or hf(x).
So g'(x) is equal to the limit as h→0, which is also equal to f(x). I don't fully understand this step, but I think it's because I don't really remember this method for taking derivatives. Again, I'll ask about it in class and go look that process up again later.
So we get the first part of the Fundamental Theorem of Calculus. If f is continuous on [a, b], then g(x) (which is the integral from a to x of f(t) as long as x is between a and b) is continuous and differentiable on [a, b] and, more importantly, g'(x)=f(x).
Reworded, assuming the normal constraints of continuity and whatnot, if we take the integral of a function from a to x, and define it as a function of x, then the derivative of that new function is the antiderivative of the original integrated function. As the book puts it, if we integrate f, then differentiate the result, we get back where we started.
This realization allows us to look at principles discovered with integral calculus (Like the Fresnel function, provided by the book as a function important to optics) and let us play with it in all the ways we can do with differential calculus. It's like realizing you can stick other games into Sonic and Knuckles other than just a Sonic game and play new bonus stages.
The next part of the FTC provides a much easier method of integration than chaining together all of the Riemann sums that took forever and made me stay up until all hours doing my homework.
FTC2: The integral of a function from A to B is the antiderivative of B minus the antiderivative of A. Or, if you know an antiderivative of f, then we can integrate the function by subtracting the the antiderivatives at the endpoints.
I must say...that's irritatingly simple.
So the Fundamental Theorem of Calculus is basically 2 parts. If a function is integrated then differentianted, you end up back where you started, and the integral of a function equals any of its antiderivatives at its two given endpoints. Or, integration and differentiation are inverse processes.
Saturday, January 22, 2011
5.2 - The Definite Integral (now with ALT symbols)
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.
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.
Friday, January 21, 2011
5.1 - Areas and Distances
Chapter 5 begins with Integrals, which, according to DBU at least, separates Calc 1 thinking from Calc 2 thinking. I remember ending last semster with integration, though I don't remember much of the technique behind it. So, note to self: Refresh integration technique.
So we want to know the area under a curve of function. Given the vague function y=F(x), we see the area that is encased between the origin, the curve, and within given stopping points A and B. Since the sides are all curvy, it's not as quick to figure out as straight shapes like squares, triangles, etc. So we have to come up a precise, mathy definition of what an area means.
We figured out the definition of a tangent line by taking two points - the point we wanted the tangent of, and another point - and moving that second point closer and closer to our tangent point. The slope would keep approaching the slope of what the tangent line would be, so the limit of the slope was the slope of our tangent line. We're going to do pretty much the same thing to get the area under the curve, but with rectangles.
Example
We're given the curve y=x^2, and want to know the area under in from 0 to 1.
By making vertical lines from an X value every 1/4 distance (1/4, 1/2, 3/4, 1) we cut the area into four rectangles. We decided the width was going to be 1/2, and, since the length is how far from the origin to the line, the length will be f(x) for whatever point we're on.
We can extend these lines into full rectangles and add the area of all of them, which, in the example, comes out to .46875. We know that's too much, so we cut the rectangles' widths in half, doubling the number of them. We have to decide whether to use f(x) as the left end of the rectangles or the right end, which results in us either overshooting or undershooting our answer.
The smaller we make the rectangles, the closer we get to the actual area of the curve. Just looking at the numbers from the example, it looks like we keep getting closer and closer to 1/3 as our area.
Example 2
So let's see.
We're going to call Rn (there's not really a convenient way to write subscripts in basic HTML. Oh well.) the sum of all of the areas of the rectangles in our area. If n is the number of rectangles we're using, then the area of each will be (1/n)x(1/n)^2 + (1/n)(2/n)^2 since (1/n) is the width we chose, and we're increasing it by 1 and squaring it each time to get the length.
We apparently have a formula for how to summarize the sum of the squares of first n positive integers, mentioned casually as if I'm supposed to hold every formula I've seen since Algebra 1 in my head forever. I'll look that one up later, and take their word for it now. Anyway, this formula is [n(n+1)(2n+1)/6] Multiplying that by the 1/n^3 we pulled out of our series, we get [(n+1)(2n+1)/6n^2].
While taking the limit of that, we can separate it into three separate fractions: (1/6)[1+1/n][2+1/n]. If we let n go to infinity, we see the limit becomes 1/3.
Since the limit of the sum of the areas of rectangles inside this curve is 1/3, we know that the area of the space under the curve is 1/3. Yipee!
So we go back to our weird curvy thing at the beginning of the chapter. To get general principles from it, we're labeling everything with variables. So we take S, the area under the curve, and divide into n strips, which we label Sv1 (little V is a subscript now. It's like squaring^, but pointing down. Deal with it.), Sv2, all the way to Svn, and assume they will all have equal length because we don't need to make it hard on ourselves unless we're looking for something fun to present a paper on.
a and b are the endpoints of our interval, so if we're dividing it up into n equal segments, the width of these will be (b-a)/n, or (Delta)x. Since we're starting at a, the actual points of the rectangle segments (assuming we're using them as the right ends) are xv1 = a+(del)x, xv2 = a+2(del)x, etc.
So the sum of all of these rectangles will be the width(which is our delta X) times the hight (which is the value of the function at xvn.) and we substitute the values of whatever problem we're working on.
Since the estimate gets better the more rectangles we have (the higher the value of n), we can define the area under a curve as the limit of Rvn as n approaches infinity. It also works for the left endpoints, so Lvn also works. It then shows that we can say "screw endpoints!" altogether and start at any sample point, and the theorem still works. The symbol for "any sample point" is xvi with a star above the i. I wonder how long it will be before we run out of symbols completely.
Since that takes up a lot of room, we can use the sigma notation of (sigma from i=1 to n) [f(xvi)](Del)x. Much better.
---
The next example just work out a problem, so I'm not summarizing that. Next is the Distance Problem.
We want to know how far something went in a given time if its velocity kept changing, like someone texting while driving. If you graphed it out, you'd get a curvy graph like the ones we just figured out the area of.
Example 4
A car's odometer is broken and the poor OCD math major who owns the car can't get it fixed yet, but has to know exactly how far he went. He's not car-savy enough to fix it himself, the shop kids still give him wedgies, but he is smart enough to get a device running to capture the exact speed he was going every five seconds.
I'm not recreating the table here, but every 5 seconds we get speed readings. Take my word on the values.
Seeing that the velocity doesn't change drastically during the 5-second intervals, we're going to pretend we were going exactly as fast as the first interval says for the whole 5 seconds. So if he went 25 ft/sec for the first five seconds, he covered 125 feet. The next interval has him going 31 ft/sec for the next 5 seconds, so that adds 155 feet. If we do this for the whole table, we end up with an estimate of 1135 feet. We can also do this using the speed at the end of the 5 second period instead of the speed at the beginning, which would give us an estimate of 1215 feet.
Lo and behold, we are actually taking equal distances on the x axis, and multiplying them by their value where they meet the graph we'd make from the points. This sounds suspiciously like what we were just doing. Sure enough, if we plot the points we figured out and make a rectangle out of the y values, it looks just like what we were doing. The area of the rectangle under a segment is roughly the distance traveled.
So we can re-write our "area-under-the-curve" formula to be a "distance-traveled-by-weird-speeds" formula by making (del)(t) instead of (del)(x)and f(tvi) instead of f(x). Since measuring the time more frequently makes smaller, and thus more, rectangles, the limit of the above equation as n approaches infinity is the distance travelled by the thing we're tracking. Or: The distance traveled is equal to the area under its velocity function.
The book assures us that the area under a curve ends up representing all kinds of fun things later on.
Meanwhile, the nerd with the broken odometer, if he managed to avoid careening into oncoming traffic while figuring this out, completely geeks out, pulls over to the side of the road, writes an Android application to do these calculations and replaces his odometer with it, resulting in him being late for the first date he'd gotten in years.
So we want to know the area under a curve of function. Given the vague function y=F(x), we see the area that is encased between the origin, the curve, and within given stopping points A and B. Since the sides are all curvy, it's not as quick to figure out as straight shapes like squares, triangles, etc. So we have to come up a precise, mathy definition of what an area means.
We figured out the definition of a tangent line by taking two points - the point we wanted the tangent of, and another point - and moving that second point closer and closer to our tangent point. The slope would keep approaching the slope of what the tangent line would be, so the limit of the slope was the slope of our tangent line. We're going to do pretty much the same thing to get the area under the curve, but with rectangles.
Example
We're given the curve y=x^2, and want to know the area under in from 0 to 1.
By making vertical lines from an X value every 1/4 distance (1/4, 1/2, 3/4, 1) we cut the area into four rectangles. We decided the width was going to be 1/2, and, since the length is how far from the origin to the line, the length will be f(x) for whatever point we're on.
We can extend these lines into full rectangles and add the area of all of them, which, in the example, comes out to .46875. We know that's too much, so we cut the rectangles' widths in half, doubling the number of them. We have to decide whether to use f(x) as the left end of the rectangles or the right end, which results in us either overshooting or undershooting our answer.
The smaller we make the rectangles, the closer we get to the actual area of the curve. Just looking at the numbers from the example, it looks like we keep getting closer and closer to 1/3 as our area.
Example 2
So let's see.
We're going to call Rn (there's not really a convenient way to write subscripts in basic HTML. Oh well.) the sum of all of the areas of the rectangles in our area. If n is the number of rectangles we're using, then the area of each will be (1/n)x(1/n)^2 + (1/n)(2/n)^2 since (1/n) is the width we chose, and we're increasing it by 1 and squaring it each time to get the length.
We apparently have a formula for how to summarize the sum of the squares of first n positive integers, mentioned casually as if I'm supposed to hold every formula I've seen since Algebra 1 in my head forever. I'll look that one up later, and take their word for it now. Anyway, this formula is [n(n+1)(2n+1)/6] Multiplying that by the 1/n^3 we pulled out of our series, we get [(n+1)(2n+1)/6n^2].
While taking the limit of that, we can separate it into three separate fractions: (1/6)[1+1/n][2+1/n]. If we let n go to infinity, we see the limit becomes 1/3.
Since the limit of the sum of the areas of rectangles inside this curve is 1/3, we know that the area of the space under the curve is 1/3. Yipee!
So we go back to our weird curvy thing at the beginning of the chapter. To get general principles from it, we're labeling everything with variables. So we take S, the area under the curve, and divide into n strips, which we label Sv1 (little V is a subscript now. It's like squaring^, but pointing down. Deal with it.), Sv2, all the way to Svn, and assume they will all have equal length because we don't need to make it hard on ourselves unless we're looking for something fun to present a paper on.
a and b are the endpoints of our interval, so if we're dividing it up into n equal segments, the width of these will be (b-a)/n, or (Delta)x. Since we're starting at a, the actual points of the rectangle segments (assuming we're using them as the right ends) are xv1 = a+(del)x, xv2 = a+2(del)x, etc.
So the sum of all of these rectangles will be the width(which is our delta X) times the hight (which is the value of the function at xvn.) and we substitute the values of whatever problem we're working on.
Since the estimate gets better the more rectangles we have (the higher the value of n), we can define the area under a curve as the limit of Rvn as n approaches infinity. It also works for the left endpoints, so Lvn also works. It then shows that we can say "screw endpoints!" altogether and start at any sample point, and the theorem still works. The symbol for "any sample point" is xvi with a star above the i. I wonder how long it will be before we run out of symbols completely.
Since that takes up a lot of room, we can use the sigma notation of (sigma from i=1 to n) [f(xvi)](Del)x. Much better.
---
The next example just work out a problem, so I'm not summarizing that. Next is the Distance Problem.
We want to know how far something went in a given time if its velocity kept changing, like someone texting while driving. If you graphed it out, you'd get a curvy graph like the ones we just figured out the area of.
Example 4
A car's odometer is broken and the poor OCD math major who owns the car can't get it fixed yet, but has to know exactly how far he went. He's not car-savy enough to fix it himself, the shop kids still give him wedgies, but he is smart enough to get a device running to capture the exact speed he was going every five seconds.
I'm not recreating the table here, but every 5 seconds we get speed readings. Take my word on the values.
Seeing that the velocity doesn't change drastically during the 5-second intervals, we're going to pretend we were going exactly as fast as the first interval says for the whole 5 seconds. So if he went 25 ft/sec for the first five seconds, he covered 125 feet. The next interval has him going 31 ft/sec for the next 5 seconds, so that adds 155 feet. If we do this for the whole table, we end up with an estimate of 1135 feet. We can also do this using the speed at the end of the 5 second period instead of the speed at the beginning, which would give us an estimate of 1215 feet.
Lo and behold, we are actually taking equal distances on the x axis, and multiplying them by their value where they meet the graph we'd make from the points. This sounds suspiciously like what we were just doing. Sure enough, if we plot the points we figured out and make a rectangle out of the y values, it looks just like what we were doing. The area of the rectangle under a segment is roughly the distance traveled.
So we can re-write our "area-under-the-curve" formula to be a "distance-traveled-by-weird-speeds" formula by making (del)(t) instead of (del)(x)and f(tvi) instead of f(x). Since measuring the time more frequently makes smaller, and thus more, rectangles, the limit of the above equation as n approaches infinity is the distance travelled by the thing we're tracking. Or: The distance traveled is equal to the area under its velocity function.
The book assures us that the area under a curve ends up representing all kinds of fun things later on.
Meanwhile, the nerd with the broken odometer, if he managed to avoid careening into oncoming traffic while figuring this out, completely geeks out, pulls over to the side of the road, writes an Android application to do these calculations and replaces his odometer with it, resulting in him being late for the first date he'd gotten in years.
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