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Lecture

6.6 FTC.pdf

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Department
Mathematics
Course
MAT237Y1
Professor
Dan Dolderman
Semester
Fall

Description
Calc 1 Tues 11/18/08 6.6 - The Fundamental Theorem of Calculus Theorem: Fundamental Theorem of Calculus This theorem is so important it had to be split up into two parts. Here’s part 1: If f(x) is continuous on [a;b] and F(x) is an antiderivative of f(x), then Z b f(x) dx = F(b) ▯ F(a) a This says the area under any (continuous) curve is F(b) ▯ F(a). In other words, just plug in the end points and subtract and you have the area. Finally, we’re o▯ of solving for the area under a curve and getting back a function...we now get actual numbers for the area. For notation purposes, I’ll be writing this also as Z b ▯b f(x) dx = F(x)▯ ▯ a a it means the same thing as F(b) ▯ F(a). You might also see in the book this notation: Z b ib f(x) dx = F(x) a a Remember, x is a variable here, while a and b are constants. Example: Z ▯ 3 2▯3 2 2 4 ▯ 2x dx = 4x ▯ x ▯ = 4(3) ▯ 3 ▯ [4(▯1) ▯ (▯1) ] = 12 ▯ 9 ▯ [▯5] = 8 ▯1 ▯1 Compare this with using geometric formulas if you don’t believe me (although why wouldn’t you? This is a sweet shortcut! Why do the extra work?)! So, the idea here is that you antidi▯erentiate exactly as you would in 6.2 and then plug in the two numbers to get out the actual area. Example: Evaluate Z ▯=2 cos(x) dx ▯▯=2 R Recall that cos(x) dx = sin(x) + C, so this becomes ▯▯=2 sin(x)▯ = sin(▯=2) ▯ sin(▯▯=2) = 1 ▯ (▯1) = 2 ▯▯=2 Okay, so you might be thinking right now: what happened to that +C? How can we just ignore it when it’s needed so much when we do inde▯nite integrals? Here’s why: Say you have a function f(x) nice and continuous and its antiderivative is F(x) + C. This means that according to FTC part 1, Z b ▯b f(x) dx = F(x) + C ▯ a a 1 so, plug in a and b, and we get F(b) + C ▯ [F(a) + C] = F(b) + C ▯ F(a) ▯ C = F(b) ▯ F(a) Aha! The C’s just disappear! They’re no longer needed! So, it’s important to remember that with de▯nite integrals (the stu▯ with numbers and FTC) there is no +C involved; however with inde▯nite integrals (no numbers...the stu▯ from 6.2, when you get a function), you must have that +C! Remember this property? Z Z b a f(x) dx = ▯ f(x) dx a b and i told you to think about it like you would think of velocity in physics: if you go backwards, the sign is negative. Well, now that we’ve done the FTC, we can see why this is true: Z b Z a f(x) dx = F(b) ▯ F(a) = ▯[F(a) ▯ F(b)] = ▯ f(x) dx a b Warning: Notice how i said that f(x) has to be continuous in order to apply FTC. Be very careful that it is because what happens if we take a look at the function f(x) = 1 on [▯1;1]? x2 Well, there’s a VA at x = 0, but what let’s see what occurs if we ignore that minor detail and try using FTC: Z 1 ▯ 1 1 ▯ x 2 dx = ▯ x ▯1 = ▯(1 ▯ (▯1)) = ▯2 ▯1 1 Yeah, we get out a number, but it’s wrong. Why? Think about it, x2 is always positive. How could the area under it be negative? So, don’t just blindly use FTC thinking it’ll solve all your problems. Make sure that you actually can use FTC before using it. So what if a question like this were to show up on an exam? Then you tell me that FTC doesn’t apply because there’s a VA at x = 0 (ie it’s not continuous on the interval we’re dealing with) and then go on to the next problem. Let’s do a couple more examples. Example: Z 4 p 3 x dx 1 2 3=2▯ = 3 ▯ x ▯ 3 1 = 2[4 3=2▯ 1 3=] = 2[8 ▯ 1] = 14 2 Example: p Z 2=2 dx p 0 1 ▯ x ▯p ▯1 ▯ 2=2 = sin x▯0 ▯1 p ▯1 = sin ( 2=2) ▯ sin (0) ▯ = ▯ 0 4 ▯ = 4 Example: Z ▯=2 x + 2 dx ▯=6 sin (x) Z ▯=2 = x + 2csc (x) dx ▯=6 2 ▯ x ▯=2 = 2 ▯ 2cot(x)▯=6 ▯ ▯ ▯ ▯ ▯2 ▯▯▯ ▯2 ▯ ▯ = 2 ▯ 2cot ▯ 2 ▯ 2cot 2 ▯ 2 2 6 ▯ 2 6 ▯2 ▯ 2 p = ▯ 0 ▯ + 2 ▯ 3 8 72 8▯2 p = + 2 3 72 ▯2 p = + 2 3 9 Example: Z 1 1 1=22x 1 ▯1 = ln(x▯ 2 1=2 = 1 (ln(1) ▯ ln(1=2)) 2 ▯ln(1=2) = 2 ln(2) = 2 Example: Z 4 1 p p ▯ 3 t dt 1 t Z 4 = t▯1=2▯ 3t1=2dt 1 ▯4 = 2t 1=2▯ 3 ▯ t=2▯ 3 1 1=2 3=2 1=2 3=2 = 2 ▯ 4 ▯ 2 ▯ 4 ▯ [2 ▯ 1 ▯ 2 ▯ 1 ] = 4 ▯ 16 = ▯12 3 Okay, before we go on to FTC part 2, let’s take a look at a somewhat new concept: Total Area. This is di▯erent from the Net Area in the sense that I don’t want to know speci▯cally the area under the curve and whether it’s positive or negative etc. Instead we are going to look at all the area as if it were all positive. In other words, the Total area is the area under the curve jf(x)j. ie, Z b Total Area = jf(x)j dx a How do we go about calculating the total area? Well, it’s pretty much the same as if you were calculating the net area (just taking the integral like you’ve been doing all along). What you have to do di▯erently is split up the integral into the intervals that f(x) is positive and negative. On the intervals that f(x) is positive, leave it alone. However, on the intervals that f(x) is negative, make sure that you negate everything. Here’s an example to see what’s up: Example: Find the total area under f(x) = 2x over the interval [▯2;3]. The idea is that we have to observe when f(x) = 2x is positive and when it is negative. In this case, f(x) changes sign at x = 0, so we have to split up our integral like this: Z 3 Z 0 Z 3 ▯0 ▯3 j2xj dx = ▯2x dx + 2x dx = ▯x ▯ + x2▯ ▯2 ▯2 0 ▯2 0 and now we throw in the end points using FTC, which gives us ▯0 ▯ (▯(▯2) ) + 3 ▯ 0 = 4 + 9 = 13 Notice how this is di▯erent from the net area, which would be Z 3 2▯3 2 2 2x dx = x ▯ = 3 ▯ (▯2) = 9 ▯ 4 = 5: ▯2 ▯2 Let’s take a look at one more Example: ▯nd the total area under f(x) = 1▯ cos(x) over the interval [0;▯=2]. 2 Z ▯ ▯ Z Z ▯=2▯1 ▯ ▯=31 ▯=21 ▯
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