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Chapter 5: Integrals

I do have a question, but for now I just want to find out how to type the integral sign.

There we go. Now for my question:

3) Evaluate ∫<sub>0</sub><sup>1</sup> ( x + [ 1 - x<sup>2</sup> ]<sup>1/2</sup> ) dx

by interpreting it in terms of areas.

My work:

Using what I believe to be the Riemann sum I divided [0,1] into 5 seperate subintervals. Adding the areas of the subintervals together I get 1.26. The book answer is:

1/2 + π/4

Why is this?  2.

3. Well, consider what shape the function:

(1-x<sup>2</sup>)<sup>1/2</sup>

curves out for 0 ≤ x ≤ 1. Then see if you can use areas you already know to calculate this.  4. So we can take this integral apart and find that we have a triangle of height 1 and base 1 and a circle with the radius = 1. If you add the areas together you get 1/2 + π/4. On this next problem:

4) Express

lim n->∞ Σ<sub>i=1</sub><sup>n</sup> sin x<sub>i</sub> ∆x

as a definite integral on the interval [0, π] and then evaluate the integral.

My work:

I've found the integral to be
∫<sub>0</sub><sup>π</sup> sin x dx

but to find the area, would I use the Riemann sum or is there a different way to find the area. I know that the antiderivitive of sin x is -cos x and that ∫<sub>0</sub><sup>π</sup> sin x dx = -cos x + C, but it seems to me that I cant evaluate -cos x + C because I don't know what C is.  5. and by the way, how do you type a character that is described as U+2206 such as ∆ without doing copy-paste from the character map? I guess I only mean this character specifically. I know how to do the others that I have typed so far.

Me trying to figure it out:

Windows says:
If you know the Unicode equivalent of the character you want to insert, you can also insert a special character directly into a document without using Character Map. To do so, open the document and position the insertion point where you want the special character to appear. Then, with NUM LOCK on, hold down the ALT key while using the number pad keys to type the Unicode character value.
This doesn't seem to work with 2206.  6. Are you working through exercises at the end of each section, or are you only doing the exercises at the end of the chapter? The evidence points toward the latter, because this very issue ("Which antiderivative should I use when evaluating a definite integral?") implicitly comes up every time you take a definite integral, but after you do a few problems like this (for example, 19-42 in section 5.3), you realize it actually doesn't matter. Let me explain in general why it doesn't matter, though.

So the second part of the Fundamental Theorem of Calculus (FTC) says that:

∫<sub>a</sub><sup>b</sup> f(x) dx = F(b) - F(a)

where F(x) is any function satisfying F'(x) = f(x). For any constant C, the function G(x)=F(x)+C is also an antiderivative of f(x). So what happens if we compute the definite integral with G(x) instead of F(x)?

G(b) - G(a) = (F(b)+C)-(F(a)+C) = F(b) - F(a)

So this calculation is independent of the antiderivative you choose, and you never have to specify a specific constant! In fact, most people end up just thinking of, e.g., -cos(x) as the antiderivative of sin(x), even though this isn't technically true. You don't really have to start worrying about constants unless you're trying to find a specific antiderivative for some reason (e.g., finding position as a function of time given initial position and velocity as a function of time) or you start solving differential equations.

So, to answer your specific question, using FTC with -cos(x) is correct.  7. The method for integrating definite integrals is slightly different to integrating indefiniate integrals. As Serpicojr says, you don't have to worry about adding a constant, 'C' to the end of the integral.

Also you have to substitute the upper and lower bounds of your interval (in your case 0 and π ) into the integral, and subtract the result of the lower bound, from the result of the upper. Also, make sure you calculator is in radians mode when you are working with radians. You probably knew that, but better safe than sorry!

For character referencing, I just copy and paste from this page   8. You are correct serpicojr. I have been for the most part only working through the exercises at the end of the chapter. Every once in a while, when I come across something that looks like I would need some extra practice with, I'll do a few extra excerises. I did a few excercises wtih sigma notation this chapter and a couple Riemann sum problems, but usually I just read the section material. I must have been tired when I read FTC, but I understand it well enough now after doing a couple problems with it. The reason I have been skipping a lot of the section excercises is because I didn't think it took 70 problems on the same concept to understand the concept. I was doing the chapter review questions instead because of the variety of the concepts they cover. I guess I asked my question prematurely though without doing much rereading which I should have done first.

Anyway guys, here is a problem that has been difficult for me to work out:

20) Evaluate the integral, if it exists.

∫<sub>-1</sub><sup>1</sup> (sin x / [ 1 + x<sup>2</sup> ] )

My work:

I know this is a Substitution Rule question, but I can't seem to figure it out. Integrating from division is a bit difficult for me right now.

Thanks for the character referencing page. I'm using it now as well.  9. Understanding the concept is one thing. Being fluent with the calculations is another. And calculus is a subject which is borderline between the conceptual and the computational--you have to understand what it is to apply it, but you have to be able to go through the steps to apply it! And being able to go through the steps necessitates doing exercise.

In this chapter, I'd say the important things are:

-taking antiderivatives
-calculating definite integrals using FTC and antiderivatives
-the fact that definite integrals can be interpreted as areas

The things which are decently important but aren't stressed as much in this book or which come up much later are:

-sigma notation
-using FTC to take derivatives

I would say don't stress about Riemann sums. They're not computationally useful (well, okay, they can be, but we prefer not to use them), and they really should be thought of as a concept necessary for the definition of a definite integral. However, Stewart uses them quite frequently to "prove" that you can use integrals to calculate various things. Since you're teaching yourself from the book, it's a good idea to understand what they are. If I were teaching you this class, however, I'd skip Riemann sums.

In any case, do a lot of integrals, particularly definite ones (because these basically involve taking indefinite integrals)! And go back and do a lot of problems involving derivatives! You'll really appreciate it in the end!  10. Looks like you'd have to use 'integration by parts' for that problem. Not sure if that will have came up yet in the book you're using?  11. not yet, but ive heard of it. There are problems like these in the section 5.5 "The Substitution Rule."

It looks like Integration by Parts comes up in two more chapters.

7.1 Integration by parts  12. There may be a substitution that works, but if so, it's far from obvious. Integration by parts, which you don't have at your disposal yet, will probably be pretty difficult. At the end of the day, though, neither of these is the easiest way of going about this problem.

I, too, was finding the problem difficult. To get my head around it, I graphed it. And then I saw it.  13. So from the graph it seems the answer is 0, which seems a little convenient for our one problem. In order to get a more universal grasp on these types of problems I've been going over 5.5 The Substitution Rule a lot in the last day or so, but these problems seem difficult for me right now. I think what the strategy is, is to set the value of u to one part of the integral, differentiate u, and then substitute something contained within the differentiated equation for u with something inside the integral.

So I have this problem in the 5.5 Exercises:

4) Evaluate the integral by making the given substitution.

∫ [ ( sin [x<sup>1/2</sup>] ) / ( x<sup>1/2</sup> ) ] x'

u = x<sup>1/2</sup>

My work:

∫ [ ( sin u ) / u ] x'
u' = ( -x<sup>-3/2</sup> / 2 ) x'

OK. I'm lost.  14. So can you think of a good reason as to why it would be 0? And, yeah, it is just convenient for this one problem, but it shows you that sometimes it's better (or necessary) to step outside the box.

For the next one, are you letting x' mean dx? If so, this is neither notationally nor mathematically correct: x' means dx/dx, a derivative; dx is a "differential", which can be thought of as an infinitesimally small change in x. In the context of an integral, I like to say the presence of the dx reminds you that you're integrating with respect to x. The absence of a differential means that you're not stating what you're integrating with respect to, and this makes the integral meaningless.

Now when you're doing this substitution, you should write down the u and du first and then figure out which pieces of the integral can be replaced by u or du. And you have to get rid of all instances of x for the substitution to work, as x depends on u. So we have:

u = x<sup>1/2</sup>

du = (du/dx)dx

What is du/dx? (I think you miscalculated this.)  15. I knew a question like this would come up because I don't fully understand Leibniz notation yet. On page 169 in Chapter 2 the book gives examples of how you may write y', but it also implies that the expaination that the book gives here of Leibniz notation is not complete.

I have a question to understand y' in relation to the notation dy/dx then. I understand that y' is a rate of change. If y' is a rate of change then it would probably have to be in terms of one unit per another unit. So lets say that y' represents 50 miles per hour. Could then dy be units of miles and dx be units of hours? If not, how would dx/dy be related to something like 50 miles per hour?  16. In the context of this book, y' and dy/dx mean the same thing. However, the notation y' assumes that we have an independent variable which is clear from context; this is usually x or, if we're dealing with time, t. The notation dy/dx clears up this ambiguity. You should read dy/dx as "the derivative of y with respect to x".

Both of these things are very different from differentials, which the things that look like dx, dy, du, etc. In the context of this book, there are a few interpretations of these symbols which you should use:

1. dx represents an infinitesimal change in x; or
2. dx reminds us that we're integrating with respect to x in an integral.

The expression dy/dx should not be thought of as "the differential dy divided by the differential dx", even though a lot of facts which jibe with this interpretation are true. For example, when I make a substitution u = f(x), I have to figure out how the differential du relates to dx. It turns out that:

du = (du/dx)dx

This should be read as "an infinitesimal change in u is equal to an infinitesimal change in x times the derivative of u with respect to x". Even though du and dx are both infinitesimal changes, du is bigger (or smaller) than dx by a factor of du/dx.

In your question, if y' represents 50 mph, then so does dy/dx (letting x be the variable describing time). We could think of dy as being an infinitesimal change in y (i.e., an infinitely small distance) and the same for dx in terms of x (i.e., an infinitely short moment).  17. Is d/dx an opperator then like + or - or / ?

Is (d/dx) y the same as dy/dx and y'?

If dy/dx = lim ∆x->0 ( ∆y / ∆x ), why does ∆x have to approach 0? I know that dy/dx should not be reguarded as a ratio, and for the time being I'll treat it like an opperator, but wouldn't the ratio of ∆y/∆x be the same no matter what value is used for x? I guess this particular question is about why
This should be read as "an infinitesimal change in u is equal to an infinitesimal change in x times the derivative of u with respect to x".
And also, what does it mean to say "with respect to x"?  18. Is d/dx an opperator then like + or - or / ?
Pretty much, though I'm not sure whether it is actually regarded as being strictly an 'operator' between mathematicians.

Is (d/dx) y the same as dy/dx and y'?
Yeh, it means the same thing.

If dy/dx = lim ∆x->0 ( ∆y / ∆x ), why does ∆x have to approach 0? I know that dy/dx should not be reguarded as a ratio, and for the time being I'll treat it like an opperator, but wouldn't the ratio of ∆y/∆x be the same no matter what value is used for x? I guess this particular question is about why
Because as ∆x approaches 0, the ratio, ∆y / ∆x gets closer and closer to the REAL value of the gradient at that particular point. Wikis article explains it well, using graphical representation:

http://en.wikipedia.org/wiki/Derivat...the_derivative

And also, what does it mean to say "with respect to x"?
When you consider the 'rate of change' of some quantity, then it must be changing 'with respect to' something. For example we could talk about the rate of change of velocity of a particle in an electric field....but with respect to what? time? the position of the particle within the field? field strength? you must specify what the change is with respect to.  19. (double posted that^^^)  20. bit4bit did a good job of answering your questions, but let me add my two cents. Originally Posted by Demen Tolden
Is d/dx an opperator then like + or - or / ?
Yes, except d/dx only acts on functions.

Is (d/dx) y the same as dy/dx and y'?
d/dx means "take the derivative of [whatever] with respect to x", and dy/dx is the derivative of y with respect to x, so (d/dx)y is the same as dy/dx.

If dy/dx = lim ∆x->0 ( ∆y / ∆x ), why does ∆x have to approach 0?
Derivatives were originally created to represent instantaneous rates of change--e.g., the velocity of a particle at a given moment. Our notion of derivative says that we can approximate the instantaneous rate of change by average rates of change--e.g., the average velocity of a particle over a period of time. This is exactly what ∆y/∆x is--an average. Our approximation should be better as ∆x gets smaller, as the instantaneous rate of change shouldn't vary too much over a small interval. That is why we let ∆x go to 0.

...wouldn't the ratio of ∆y/∆x be the same no matter what value is used for x?
Let's think about what ∆y and ∆x mean. We're always concerned with a derivative at a point, so let's let this point be x. So we're averaging the change in y over the interval from x to x+∆x. y is a function of x, y = f(x), so our average is really:

∆y/∆x=[f(x+∆x)-f(x)]/∆x

This ratio will be constant only if f(x+∆x)-f(x) is a multiple of ∆x, and you can show this only happens if f(x) = ax+b for some real numbers a, b. So, in general, this ratio will not be constant.

And also, what does it mean to say "with respect to x"?
When I say "with respect to x", I mean that x is the independent variable, and we are studying how other variables change as x is changed. Let me show you an example that will make more sense when you do multiple variables. Imagine a function which takes two real variables x and y, say:

f(x,y) = 2x-2y

x and y are independent. Each can vary without changing the other. Thus I can ask how the function changes as x changes and y is held constant. Evidently, the instantaneous rate of change as x varies is 2. Similarly, the instantaneous rate of change as y varies is -2. I would call the first of these two values the (partial) derivative of f with respect to x, the second the same with respect to y.

I hope this helps...!  21. Ok, thanks. All of that makes sense to me I think.

So there is one particular problem in the examples of 5.5 that I have been trying to use to understand the substitution process. This is the example and beginning of the book solution:
-------------------------------------------------------------------
Example 3: Find ∫ ( x / [ ( 1 - 4x<sup>2</sup>)<sup>1/2</sup> ] ) dx

Solution: Let u = 1 - 4x<sup>2</sup>. Then du = -8x dx, so x dx = -1/8 du and...
-------------------------------------------------------------------

I know that du/dx = -8x, but it looks like the book is treating du/dx like a ratio by saying du = -8x dx  22. Right, this is an instance of where the du/dx acts like a ratio. The book is using the fact that:

du = (du/dx)dx

To you, this should just mean that you can't just dx by du in your integral; you have to be able to find du/dx floating around in the integral, and then you have to get rid of it to change du to dx. This is the key step in doing substitution.  23. I think I understand d/dx or dy/dx or just dx at the end of an integral. I believe all three mean that f(x) changes with respect to x. In the case of an integral, the integral really is f(x). Because I understand dx this way, not as a value, but as a context of, I don't understand how you can have something like du = -8x dx. That equation doesn't really make sense to me. The way I understand du, it shouldn't be able to equal to anything. Maybe I need to keep thinking of it in terms of the limit as x->0 so the equation really says Δu = -8x Δx as Δx->0, but then Δu->0 when Δx->0.

I was starting to get used to the idea that d/dx is not a ratio.  24. What kinds of exercises could I do to get used to what Leibniz notation means in different circumstances and what its parts mean when its taken apart? Your explanations have been helpful, but I think for me to really understand it well enough I need to have practice.  25. Originally Posted by Demen Tolden
Maybe I need to keep thinking of it in terms of the limit as x->0 so the equation really says Δu = -8x Δx as Δx->0, but then Δu->0 when Δx->0.
That's the right idea. Note that, although Δu->0 when Δx->0, the ratio Δu/Δx approaches -8x.

Equating differentials is a confusing thing at this point in calculus. Just think of it as a notational convenience which makes substitution easier. The expression:

du = (du/dx)dx

really just means that, when making a substitution, you have to remember to find the term du/dx somewhere in your integral and then get rid of it when you make the substitution (i.e., change (du/dx)dx to du).

I was starting to get used to the idea that d/dx is not a ratio.
And that's the right idea--it's an operator. I don't know if there are any good exercises which will help you understand Leibniz notation. I think you'll just have to remember that:

-d/dx (or d/dz or whatever) is an operator (differentiation with respect to x, or z, or whatever)
-dy/dx (or dw/dz or whatever) is a derivative (of y with respect to x, or w with respect to z, or whatever)
-dx is a differential and tells you what you're integrating with respect to, and you really only need to worry about this in the context of substitution (and a few other things... separable equations, arc lengths...)  26. I guess I somehow figured out how to do Substitution. Although I probably really don't understand what I am doing, I can do it, meaning that if I were to explain what my equations meant, I couldn't say. I did about 80 of the 5.5 exercises last night and this morning and now I'm pretty good at Substitution. 99% of the time I somehow know exactly what substitution to use. There were only 2 problems I couldn't figure out, but 2 out of 80 shouldn't make much of a difference.  27. Do you at least understand substitution as a technique for computing integrals (both indefinite and definite)? This is enough of an understanding for you to continue your studies.

If you want a better idea of what it is, think of substitution as being the opposite of the chain rule. Substitution basically says that the integral of a function of the form f'(g(x))g'(x) is f(g(x))+C, and this is true because the chain rule tells us the derivative of f(g(x)) is f'(g(x))g'(x). We introduce the variable u=g(x) to make things easier on our eyes, then we have the formal substitution du=g'(x)dx, and this transforms the problem to integrating f'(u).  28. Yes, and I've gotten better at it still now. I will here demonstrate my skill with this problem I found a bit difficult:

Chapter review 38 ) Evaluate the integral, if it exists:

∫<sub>0</sub><sup>4</sup> | x<sup>1/2</sup> - 1 | dx

my work:

u = x<sup>1/4</sup>
du = (1/4)x<sup>-3/4</sup> dx
(1/4)u<sup>-3</sup> = (1/4)x<sup>-3/4</sup>
du = dx / (4u<sup>3</sup>)

∫ | (x<sup>1/4</sup> + 1 )(x<sup>1/4</sup> -1 ) | dx
∫ | (u + 1 )( u - 1 ) | dx
∫ | (u + 1 )( u - 1 ) | ( 4u<sup>3</sup> / 4u<sup>3</sup> ) dx
∫ | (u + 1 )( u - 1 ) | 4u<sup>3</sup> du
4 ∫ | (u + 1 )( u - 1 ) | u<sup>3</sup> du
4 ∫ | (u<sup>2</sup> - 1 ) | u<sup>3</sup> du
4 ∫ | (u<sup>5</sup> - u<sup>3</sup> ) | du
4 ∫ | u<sup>5</sup> | du - 4 ∫ | u<sup>3</sup> | du
4 ∫ | u<sup>3</sup>u<sup>2</sup> | du - 4 ∫ | u<sup>2</sup>u | du

z = u<sup>3</sup>
dz = 3u<sup>2</sup> du

a = u<sup>2</sup>
da = 2u du

(4)(1/3) ∫ | z dz | - (4)(1/2) ∫ | a da |
(4/3)(1/2) z<sup>2</sup> - (2)(1/2) a<sup>2</sup>
(2/3)z<sup>2</sup> - a<sup>2</sup>
(2/3)u<sup>6</sup> - u<sup>4</sup>
(2/3)x<sup>6/4</sup> - x<sup>4/4</sup>
(2/3)x<sup>3/2</sup> - x
or
x((2/3)x<sup>1/2</sup> - 1)

when x = 4, x((2/3)x<sup>1/2</sup> - 1) = 4/3
when x = 0, x((2/3)x<sup>1/2</sup> - 1) = 0
4/3 - 0 = 4/3, so the value of ∫<sub>0</sub><sup>4</sup> | x<sup>1/2</sup> - 1 | dx is 4/3
----------------------------------------------------------------------------------

Now I do have a question regarding problems 43 - 48. In these problems the book asks me to find the derivitive of the function. The first of these problems is:

43) Find the derivitive of the function.

F(x) = ∫<sub>1</sub><sup>x</sup> (1 + t<sup>4</sup>)<sup>1/2</sup> dt

but wouldn't the derivitive of this function be (1 + t<sup>4</sup>)<sup>1/2</sup> dt ?  29. Actually I see a mistake I made on problem 38. I should have broken the integral apart at x = 1 into two separate integrals to account for the portion that is negative.  30. You're right about 38--you should break it up into two integrals. However, you shouldn't need to use substitution to solve the problem--everything is a power of x (or constant), and these we can integrate without any special techniques.

For 43, you have the right idea--the first statement of the FTC tells you what happens when you take the derivative of such a function. But be careful with your variables, and also note that a function shouldn't have a differential in it (unless the function is defined by an integral).  31. I think I've gotten everything I need out of this chapter then. Unless one of you two wish to suggest something else I'll move on to Chapter 6: Applications of Integration. I know you, serpicojr said:
In any case, do a lot of integrals, particularly definite ones (because these basically involve taking indefinite integrals)! And go back and do a lot of problems involving derivatives! You'll really appreciate it in the end!
I've done a lot of integrals now, and I do like substitution now. Anyway, I've done probably about 150 integral problems in the last 2 days. I could do some extra derivitive problems, but I'm not sure which one's to do. I think I can do derivitives pretty well.

I don't know. Move on? Go back and review? Which should I do?  32. I'm glad to hear you've been doing a fair number of exercises! You can probably just move on to the next chapter, although don't go overboard!  33. That's awesome! You once said
I am very familiar with this book--I've taught calculus II out of it a couple of times (chs. 6-11).
So part of my motivation to get this far is to finish what would be considered Calculus I, not that I have any intention of pausing even for a day unless there is some calculus idea I have to run through my head for a while before moving on.

I guess I am just glad to have gone this far. I already have a chapter 6 question though, and I'll post it in the new thread.  34. Yeah, you've just completed what a first semester calculus class covers at my school! Great job.  Bookmarks
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