Integrals (u-substitution)

  • #26
Char. Limit
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Well, that's mainly because as the number of "rectangles" increases (or goes to infinity), the difference between the area of all those rectangles and the area under the function will tend to 0, regardless of whether you use a left, right, or a mid point.

At least I think so.

EDIT: I don't understand...? I just type in dx in the latex and it gives me what i want. Unless you mean [itex]\operatorname{d}x[/itex], which is given by \operatorname{d}x.
 
  • #27
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I'm still reading through the examples that use n=10 and other smaller numbers to do the rieman sum by hand.

Char wrote:
[tex]\int_a^b f(x) dx = F(b) - F(a)[/tex]

That is, F(b)-F(a), where F(x) = the indefinite integral of f(x)? I have a hard time grasping how the area under the curve relates to the antiderivative. It seems like that constant would leave a huge margin for error. Idk, I'll get to that part I guess.

\dif is a command for the differential operator. It is simply an upface d to
clarify that it is an operator. E.g.:
\dif x dx
from: http://www.tug.org/texlive/Contents/live/texmf-dist/doc/latex/commath/commath.pdf
 
  • #28
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Still struggling with understanding how using mid/left/right points will yield the same limit.

This is not obvious to prove in a rigorous way. The proof of this uses things like uniform continuity of the integral and Cauchy sequences. Too much for a first encounter. I suggest you just accept this and go on.

Alternatively, how do I make the roman "d" as in "dx" in latex? \diff does not appear to work?

I think the standard is to use \mathrm{d}x.
 
  • #29
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I'm still reading through the examples that use n=10 and other smaller numbers to do the rieman sum by hand.

Char wrote:
[tex]\int_a^b f(x) dx = F(b) - F(a)[/tex]

That is, F(b)-F(a), where F(x) = the indefinite integral of f(x)? I have a hard time grasping how the area under the curve relates to the antiderivative. It seems like that constant would leave a huge margin for error. Idk, I'll get to that part I guess.

That the antiderivative relates to the area is one of the most beautiful and surprising results from mathematics. I can't give you an intuitive reason why this is the case, you just need to read the proof and convince yourself from that fact. It's quite a short proof if I remember well.
 
  • #30
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I suggest you just accept this and go on.

I already have, I was just thinking that the book would explain it, at least the idea, but it looks like it's jumping right into some sample definite integral problems instead. Ah well. I don't even know what a cauchy is :rofl:

I can't give you an intuitive reason why this is the case, you just need to read the proof and convince yourself from that fact. It's quite a short proof if I remember well.

I will do that.
 
  • #31
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I already have, I was just thinking that the book would explain it, at least the idea, but it looks like it's jumping right into some sample definite integral problems instead. Ah well. I don't even know what a cauchy is :rofl:

There are two ways to learn calculus: intuitive and easy, or rigorous with proofs. You need to know it intuitively before you can do it rigorously. But that means that the books will leave out facts that you don't find obvious at first. You'll need a more formal book if you want to understand everything , but I wouldn't advise this when first confronted with definite integrals :smile:
 
  • #32
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I'm reading from the Stewart book, and using the Spivak book as a reference, because I find it hard to understand sometimes.

Well, that's mainly because as the number of "rectangles" increases (or goes to infinity), the difference between the area of all those rectangles and the area under the function will tend to 0, regardless of whether you use a left, right, or a mid point.

At least I think so.

Now that I think about it, this makes sense, because as the number of "n" increases, the width of each "n" decreases, and since [itex]x_i[/itex] is always within [itex]\Delta x[/itex], the squeeze theorem would dictate that as [itex]\Delta x[/itex] approaches 0, so does [itex]x_i[/itex]. Does that make sense?
 
  • #33
Char. Limit
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I'm reading from the Stewart book, and using the Spivak book as a reference, because I find it hard to understand sometimes.



Now that I think about it, this makes sense, because as the number of "n" increases, the width of each "n" decreases, and since [itex]x_i[/itex] is always within [itex]\Delta x[/itex], the squeeze theorem would dictate that as [itex]\Delta x[/itex] approaches 0, so does [itex]x_i[/itex]. Does that make sense?

That makes sense to me, and is probably a good intuitive basis for this fact. Now, the Fundamental Theorem of Calculus, that doesn't really have much of an intuitive basis, unfortunately.
 
  • #34
vela
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That makes sense to me, and is probably a good intuitive basis for this fact. Now, the Fundamental Theorem of Calculus, that doesn't really have much of an intuitive basis, unfortunately.
It sort of does, actually, but I've never found it all that helpful. Say you have a function f(x). At x=x0, it has a value f(x0). You can approximate the value of the function at a nearby point x1=x0+Δx as
[tex]f(x_1)=f(x_0+\Delta x) \approx f(x_0) + f'(x_0) \Delta x[/tex]
You can see this easily from the graph. The intuition is simply that f'(x0) tells you which way the function is going at x=x0. If you know where you are, i.e., f(x0), and which way you're going, i.e., f'(x0), you can figure out where you'll end up, i.e., f(x1). If you rearrange it slightly, you already see an inkling of the fundamental theorem:
[tex]f(x_1) - f(x_0) \approx f'(x_0)\Delta x[/tex]

Repeating the process, at the point x2=x1+Δx, you obtain
[tex]f(x_2)=f(x_1+\Delta x) \approx f(x_1) + f'(x_1) \Delta x \approx f(x_0) + f'(x_0) \Delta x + f'(x_1) \Delta x[/tex]
and so on. Now let x0=a, xn=b, and Δx=(b-a)/n. Then xi=x0+i Δx, and
[tex]f(b) \approx f(a) + f'(a) \Delta x + f'(x_1) \Delta x + \cdots + f'(x_{n-1})\Delta x[/tex]
or
[tex]f(b)-f(a) \approx f'(a) \Delta x + f'(x_1) \Delta x + \cdots + f'(x_{n-1})\Delta x[/tex]
The righthand side is a Riemann sum, and in the limit Δx→0, you get
[tex]f(b)-f(a) = \int_a^b f'(x)\,dx[/tex]
 
  • #35
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I found a video of this rather strange looking man doing the proof:

Going to check to see if it's proven on ocw.mit.
 
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  • #36
Ray Vickson
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Related to this question:

[tex]\int sec^{3}(x)tan(x) dx[/tex]

[itex]u=sec(x)[/itex], and so, [itex]\frac{du}{dx}=sec(x)tan(x)[/itex], and [tex]dx=\frac{du}{sec(x)tan(x)}[/tex]

[tex]\int u^{3}tan(x) dx[/tex]
[tex]\int \frac{u^{3}tan(x)}{sec(x)tan(x)} du[/tex]
[tex]\int \frac{u^{3}}{sec(x)} du[/tex]

I have no idea what to do with this one? Can I put my u=sec back in and try again now?

Edit: Na, that just gets me back to where I started.

Oh wait, because u is equal to sec, can I just call that [itex]\frac{u^{3}}{u}[/itex] ?

You could note that [itex] f(x) = \sec^{3}(x) \tan(x)[/itex] has the form [itex] f(x) = \sin(x)/ \cos^{4}(x) [/itex], so substituting [itex] u = \cos(x) [/itex] gives [itex] f(x) dx = - du/u^4 [/itex].

RGV
 

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