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Lecture

6.4 defn area.pdf

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Department
Mathematics
Course
MAT237Y1
Professor
Dan Dolderman
Semester
Fall

Description
Calc 1 Fri 11/10/08 6.4 - Defn of Area; Sigma Notation When I explained section 6.1, I breie y talked about approximating area under curves using rectangles. The problem is, if we want a lot of rectangles to get the area a pretty good approximation, we need to add up a whole bunch of stu▯ and it got really tedious to write + over and over. Well, rather than writing something plus something plus something a whole bunch of times, mathematicians came up with a shortcut. This is capital sigma: ▯. It means sum. ie, if i were to say 1 + 2 + 3 + 4 + 5 + ▯▯▯ + 10 this is the same as expressing it in terms of ▯: X10 k k=1 Have you ever taken a computer science or a programming class before? This works the same way as a loop does. It says start at k = 1 and keep incrementing k by one until you get to 10. Oh, and by the way, add up all the k’s while you’re doing that. Notice this does not actually solve the adding problem. It merely rewrites it so that it’s less to write (imagine if i were to ask you to add up all the numbers from 1 to 100! Writing it down in this ▯ notation is much easier than writing 1 + 2 + ▯▯▯ + 100!). We shall get to tricks involving what to do about actually adding the numbers in a bit. For now, we’re just interested in the actual notation. We’re also not limited to just adding the numbers between 1 and 10 either. We can do other things. Say i wanted to add all the squares from 1 to 100. This formula would change slightly to X0 k2 k=1 It expands out to 1 + 4 + 9 + 16 + ▯▯▯ + 81 + 100. ▯ by itself just means add stu▯ up. The k = 1 on the bottom says we’re dealing with the variable k here, and it’s starting at 1. i just as well could have started at 3 if i wanted to. Finally, the 10 on top means when i have to stop adding. When i reach k = 10, stop. That’s the last one. to the right of ▯ there can be any function of variable k (say, f(k)). You don’t have to have k in that function, of course. 5 X 1 = 1 + 1 + 1 + 1 + 1 = 5 k=1 That just sums up the number 1 ▯ve times. other examples: X6 2k = 2(3) + 2(4) + 2(5) + 2(6) k=3 X0 k = (▯4) + (▯3) + (▯2) + (▯1) + (0) 3 k=▯4 1 You can also change the limits of these sums. To do this, make sure that when you plug in that ▯rst number you get what you’re supposed to get. Example: we have 1 + 4 + 9 + 16 + 25. We can express this as 5 X k2 k=1 or perhaps 4 X 2 (k + 1) k=0 Notice that when we plug in k = 0, we get 1 ; k = 1 gives 2 ; up to k = 4 which gives 5 = 25. Now, sums have a few nice properties that i think i should mention. Namely, if you’re adding a bunch of stu▯ together that has a common factor, you can take out that factor. ie, n n X X cak= c ak k=1 k=1 where k is our function (i might switch it o▯ betweek f(k) and a , but they’re essentially the same thing) and c is just a constant. n n n X X X (k + k ) = ak+ bk k=1 k=1 k=1 n n n X X X (k ▯ k ) = ak▯ bk k=1 k=1 k=1 Now that we got some of those properties out of the way, here are three summations that you should become familiar with. In fact, you should become so familiar with them that you should memorize them! n X n(n + 1) k = 1 + 2 + ▯▯▯ + n = k=1 2 Xn k = 1 + 2 + 3 + ▯▯▯ + n =n(n + 1)(2n + 1) 6 k=1 ▯ ▯ Xn n(n + 1)2 k = 1 + 2 + ▯▯▯ + n = 2 k=1 These play a crucial role in ▯nding areas because you don’t actually have to add a whole bunch of stu▯ up! You can quickly multiply a couple numbers together and voila, you’re already done! 2 Okay, so now let’s use this new knowledge and apply it to the area stu▯. As you may recall, when an interval [a;b] is split into n equal subintervals, we found that the width of each subinterval was b▯a. The b ▯ a part is the total length of the interval, and we want to divide that entire length up n n times. Well, to make things a little more convenient (and more confusing, because what would calculus be without the confusion?), we’re just going to call that ▯x. We can of course evaluate the rectangle’s height at any point within each interval we want. These points that will determine each rectangle’s height are going t1 be2x ;x n:::;x . This means that ▯ ▯ the heights themselves are f1x );:::;fnx ). ▯ So, the ▯rst rectangle would be of width ▯x and have height 1(x ), while the second rectangle has width ▯x and height f(x ), and so on. This gives us the total area of all the rectangles to be ▯ ▯ 2 ▯ ▯xf(x 1 + ▯xf(x )2+ ▯▯▯ + ▯xf(x )n Remember, we’re adding the areas of all the rectangles together. This shortens to n X ▯ A ▯ f(xk)▯x k=1 As I mentioned before, the more rectangles you do, the better approximation you have. So, take the limit of the number of subintervals (ie, the number of rectangles!), and you will get not an approximation, but the exact area. X n A = lim f(xk)▯x n!1 k=1 Also, in your homework, you’ll be approximating areas using right endpoints, left endpoints, and midpoints. I am just going to do right endpoints, because it’s basically all the same thing, just a ▯ di▯erent xkis used, that’s all. But, here they all are: ▯ Left: k = a + (k ▯ 1)▯x ▯ Right: xk= a + k▯x ▯ 1▯ Mid: x▯ = a + k ▯ ▯x k 2 3 Example: use right endpoints to ▯nd the area under f(x) = x over [0;1]. Solution: ▯rst, let’s see what ▯x is. ▯x = b ▯ a= 1 ▯ 0= 1 n n n and since we’re doing right end points, our x ’s are going to be k ▯ k k
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