# Homework Help: My nightly integral thread:

1. Jul 7, 2011

### QuarkCharmer

1. The problem statement, all variables and given/known data
I'm sitting down and doing somewhere around 50 problems this evening for practice, and I just got stuck. I will probably use this one thread to post multiple questions if that is okay.

2. Relevant equations

3. The attempt at a solution
Edited out this problem, it was my mistake.

Edit: Oh man, I see my mistake..... ( I wrote the problem down wrong)

Last edited: Jul 7, 2011
2. Jul 7, 2011

### SammyS

Staff Emeritus
Where you given a hint to use that substitution?

If you were integrating x3(2+x4)5, then that would be a good substitution.

3. Jul 7, 2011

### QuarkCharmer

Yeah, that was how it was supposed to be.

I u-sub the 2+x^4 and then the dx = du/4x^3 cancels out the other x^3. I read the problem wrong and didn't bother to recheck it lol. Yes it told me to use that substitution. I figured that one out, thank you.

4. Jul 7, 2011

### SammyS

Staff Emeritus
I started a post pursuing the m = u-2 idea, just to see, but it quickly became a nightmare.

LOL !

5. Jul 7, 2011

### QuarkCharmer

Yeah, I found that out on my own, then I glanced over and "Oh I wrote a wrong exponent" lol.

Now I am messing with this one:

$$\int sec(2x)tan(2x)dx$$

I'm thinking that I should re-write this one in terms of sine/cosine so that:

$$\int \frac{sin(2x)}{cos^{2}(2x)}dx$$
let u=cos(2x)
$$\int \frac{sin(2x)}{u^{2}(-sin(2x))}du$$
$$- \int \frac{1}{u^{2}}du$$
$$- \int u^{-2} du$$
and so on.

edit: I should have just realized that the antiderivative of sectan is sec. Duh!

Last edited: Jul 7, 2011
6. Jul 7, 2011

### SammyS

Staff Emeritus
Just use u=2x.

What is the derivative of sec(θ) ?

Otherwise, let u = cos(2x), that will work too.

7. Jul 7, 2011

### QuarkCharmer

sec(θ)tan(θ),

I see there wasn't really any work involved in this one lol, but I used the u-substitution anyway without thinking.

8. Jul 7, 2011

### SammyS

Staff Emeritus
Sorry, I kept editing my previous post.

Once you change to sines & cosines, use u = cos(2x).

9. Jul 7, 2011

### QuarkCharmer

Okay, would someone mind checking this one please?

$$\int x^{3} \sqrt{x^{2}+1} dx$$

I let $u=x^{2}+1$, such that:

$$\int \frac{x^{3}u^{\frac{1}{2}}}{2x}du$$
$$\int \frac{x^{2}u^{\frac{1}{2}}}{2}du$$

And so, $x^{2} = u-1$, such that:
$$\int \frac{(u-1)u^{\frac{1}{2}}}{2}du$$

Which simplifies to:
$$\frac{1}{2} \int u^{\frac{3}{2}} du - \frac{1}{2} \int u^{\frac{1}{2}} du$$

Taking the indefinite integral yields:

$$\frac{1}{2}(\frac{2u^{\frac{5}{2}}}{3}-\frac{2u^{\frac{3}{2}}}{3})$$

I believe the solution to be the following:
$$\frac{\sqrt{(x^{2}+1)^{5}}}{5} - \frac{\sqrt{(x^{2}+1)^{3}}}{3} + C$$
Is it correct, and should I simplify more somehow?

10. Jul 7, 2011

### SammyS

Staff Emeritus
You missed a factor of 1/2.

11. Jul 7, 2011

### QuarkCharmer

Ah, I see what you are saying. I do that alot in differentiation, I forget to that they are composition functions and don't chain-rule out the argument. I'll re-do that problem, thank you for spotting that! In that case, which should I substitute U for? The argument? u=2x ?

12. Jul 7, 2011

### SammyS

Staff Emeritus
A nice thing about finding an indefinite integral (anti-derivative) is that you can check you answer by taking the derivative of the result.

Not much of a way to simplify this. You could factor out (1/15)√[(x2+1)3], but why?

13. Jul 7, 2011

### SammyS

Staff Emeritus
When you said that u = cos(2x), then you should get du = -2sin(2x) dx.

14. Jul 8, 2011

### QuarkCharmer

Thank you sammy, I corrected the above mistake. I am working through some definite integrals at the moment, and I may want to post some questions (I have one about constants and the fundamental theorem), but I am not sure how to make a specific symbol in latex.

How do you make this symbol (in latex), and what is it actually called?

Sorry, I'm trying to use my constant PF homework help questions to my advantage by learning latex as I go, and I could not figure out how to find this symbol on the internet, and it's not in the Advanced Edit drop-downs.
Thank you.

Last edited: Jul 8, 2011
15. Jul 8, 2011

### rock.freak667

It should be \right|. It is in the latex reference button under boundaries I believe.

16. Jul 8, 2011

### I like Serena

I think we simply call it "vertical bar" and you can type it as a vertical bar: |. :)

If you follow rf's suggestion, it will come out bigger, which looks nicer.
But you will have to put \left. on the left side of your expresion to get proper latex.

To get the boundaries, you'll type a subscript/superscript combination.
Like: \right|_a^b.

17. Jul 8, 2011

### QuarkCharmer

I understand how to do the sub/super script for it. What do you mean when you say "I have to put a \left|" on the left of my equation. Can you please show me an example?

18. Jul 8, 2011

### micromass

\left. f(x) \right|_a^b

gives

$$\left. f(x)\right|_a^b$$

19. Jul 8, 2011

### QuarkCharmer

Thank you, that example helped.
Why is one \left. (period) and the other bound \left| (pipe) ?

Anyway, here is my question for tonight. I have thought about it all day for the most part. I learned about the riemann sum definition of a definite integral, did a bunch of practice problems on the subject, and watched MIT and Khan academy videos on the topic. I am finally satisfied that taking the middle/left/right to be the height of the rectangle makes no difference as n approaches infinity (squeeze theorem et al.).

Then they throw this one on me.

$$\int_{a}^{b}f(x)dx = \left. F(x) \right|_{a}^{b} = F(b)-F(a)$$

What? How the heck? This one is simply not intuitive at all for me at all.

$$F(x)=\int_{a}^{x}f(t)dt$$
$$F(x+\Delta x)=\int_{a}^{x+\Delta x}f(t)dt$$
Okay sure.

$$F(x+\Delta x) - F(x) = \int_{a}^{x+\Delta x}f(t)dt - \int_{a}^{x+\Delta x}f(t)dt$$

then they claim:
"It can be shown that"
$$\int_{a}^{x}f(t)dt + \int_{x}^{x+\Delta x}f(t)dt = \int_{a}^{x+\Delta x}f(t)dt$$
(The sum of the areas of two adjacent regions is equal to the area of both regions combined.)

Okay, here is where it breaks down for me. How are they determining that these integrals are the area? I understand that the sum of the areas from the start to the mid point, and from the midpoint to the end, is equal to the area from the start to the end. How does the integral mean area exactly?? I must have missed the part (not really) where they show how the riemann sum definition relates to the integral? I thought that is what they were basically proving here??

20. Jul 8, 2011

### micromass

For every \right there must be a left. Usually, we do something like

\left( x^2+x \right)

which yields

$$\left(x^2+x\right)$$

But here we don't want any left-bracket, we want no symbol. That is what \left. is for. So

\left. x^2+x \right)

is

$$\left. x^2+x\right)$$

You see, there is no left bracket there, there is nothing. So, if we want to type

$$\left. f(x)\right|_a^b$$

we can't just type f(x)\right_a^b since there must be a left delimiter. So we type

\left. f(x)\right|_a^b

But we know that

$$\int_a^b{f(x)dx}$$

denotes the (oriented) area under f from a to b. This was the point of the entire Riemann-sum definition. The integral is supposed to be the area!!!!

21. Jul 8, 2011

### QuarkCharmer

Okay, perhaps I just didn't see the connection between that then. I will re-read that whole proof.

I am not seeing the connection between:
$$\int_{a}^{b}f(x)dx = \lim_{n \to \infty} \Sigma_{i=1}^{n}f(x_{i})\Delta x$$
where:
$\Delta x = \frac{b-a}{n}, x_{i}=a+i\Delta x$

How does the antiderivative at one point minus the antiderivative at another point yield the area?

Last edited: Jul 8, 2011
22. Jul 8, 2011

### micromass

Check out http://www.geogebra.org/en/upload/files/UC_MAT 2009/Mary Liz Lamb/Riemann_Sums2.html

The point is that the sum $f(x_i)\Delta x$ is the area of one rectangle in the above applet. The sum $\sum_{i=1}^n{f(x_i)\Delta x}$ is the area of all the rectangles combined. And how finer we take the rectangles (and thus the more rectangles we take), the better the area of all the rectangles approximates the area under the function!

So, as you can see on the applet, if we take n very big, then the rectangles approximate the area under the function very well. So in the limit, those two areas are equal. Thus the limit of $\sum_{i=1}^n{f(x_i)\Delta x}$ is the area.

23. Jul 8, 2011

### QuarkCharmer

I understand that part, and how the limit as n tends to infinity makes the area calculated by the riemann sum. I don't get how the antiderivative relates to this though? They basically just said, "riemann sum definition = this integral notation".

/That's a neat applet.

24. Jul 8, 2011

### micromass

Ah, so you understand why $\int_a^b{f(x)dx}$ is the area?? (well, by definition)
The connection between the area and the antiderivative is the fundemental theorem of calculus. You'll need to see its proof in order to see the connection. It's not something that is easily explained...

25. Jul 8, 2011

### QuarkCharmer

Yes, I get that $\int_a^b{f(x)dx}$ is the notation so to speak of it, and that the process behind it is making infinitely small rectangles about the bounds of the integral which tend to the exact area (much in the same way a secant line tends to a tangent line in the differentiation proofs). Then they just say that "riemann sum = integral notation", well okay, I can live with that, but then when they say "Oh, hey, by the way there is an easier way of doing this than computing infinity-rectangles and it's antiderivative(b)-antiderivative(a)". Without any evidence to back it up. What is the name of the proof I am looking for then, regardless of whether or not I can understand it? "Fundamental theorem of Calculus" yields a whole bunch of the above mentioned proof (via google et al.).