Midnight Tutor - Videos To Help You Learn Calculus

1. Nov 28, 2007

jaime2000

http://www.midnighttutor.com/

I searched and it looked like no one had posted this before, so here it is. The Midnight Tutor is an awesome site in which some guys make videos of themselves in front of boards and teach calculus, be it a concept or how to solve a certain kind of problem. It is modeled on the AP Calculus program (including test specific approaches and questions), and seems to me one of the most awesome things I have seen. They also post their videos on YouTube, too. And if you have a question, you can e-mail them and they will (eventually) do a video on how to solve it! It's great! ^.^

2. Nov 29, 2007

Kurdt

Staff Emeritus
The guy knows his stuff and they are certainly useful. One criticism is that he should have planned the videos a bit more carefully as he tends to stop and start quite a lot. Never the less I'm sure lots of students would find the videos exceptionally useful.

P.S. Not too sure about his policy of solving peoples homework questions. I know he explains them thoroughly but examples would be better so the student can work through themselves.

3. Nov 29, 2007

Mk

That's an awesome site.

4. Nov 30, 2007

colby2152

I am working on something similar, but based on general Calculus lessons.

5. Dec 8, 2007

IsotropicSpinManifol

thanks for sharing!

6. Dec 9, 2007

soul

Awesome!.. This site is what I am looking for. There are many problems that I could not solve. Thanks for your sharing.

7. Dec 31, 2007

nanoWatt

Very nice. Even with my bachelors in Physics from 7 years ago I still need a refresher.

8. Jan 8, 2008

sphyics

thanks for sharing.

9. Jan 11, 2008

newbie-girl

pretty interesting
thanks for sharing the link

10. Jan 13, 2008

fermio

I found tutorial with mistake http://www.midnighttutor.com/integral_1_over_xsqrt(4x+1).html
Type 1/(x Sqrt[4x+1]) into http://integrals.wolfram.com/index.jsp
and you will see diferent
1/(1-x^2) integral in integral tables I can't found at all. only in one book found that it's equal 0.5ln|(1+x)/(1-x)|+C. But even if repair mistake wolfram reasearch still giving ln((4x+1)^0.5-1)-ln((4x+1)^0.5+1) instead -[ln(1+(4x+1)^0.5)-ln(1-(4x+1)^0.5)] if use this formula 0.5ln|(1+x)/(1-x)|+C by trying end integration calculation corectly.
btw wolfram 1/(1-x^2) integrating as 0.5(ln(x+1)-ln(x-1))=0.5ln|(x+1)/(x-1)|. So then final answer by continuing guy work must be if using wolfram formula:
-[ln((4x+1)^0.5+1)-ln((4x+1)^0.5-1)]= ln((4x+1)^0.5-1)-ln((4x+1)^0.5+1).

Last edited: Jan 13, 2008
11. Jan 13, 2008

Kurdt

Staff Emeritus
Thats the same answer.

$$\tanh^{-1}(x) = \frac{1}{2} \ln (x+1) - \ln (x-1)$$

Last edited: Jan 13, 2008
12. Jan 14, 2008

fermio

I don't think so.
arctg(1.1)=0.83298
0.5ln[(1.1+1)/(1.1-1)]=0.5ln(21)=1.52226

Final his integral
$$2 \int \frac{du}{u^2-1}$$ possible integrate by taking integral from table (which don't exist in all integral tables) $$\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\ln|\frac{x-a}{x+a}|+C$$ and then
$$2 \int \frac{du}{u^2-1}=\ln|\frac{u-1}{u+1}|+C$$

Last edited: Jan 14, 2008
13. Jan 14, 2008

Kurdt

Staff Emeritus
Its a well known identity. Besides you have calculated arctan(x) not arctanh(x). Arctanh(1.1) does not exist.

From the table you have:

$$\ln\left|\frac{u-1}{u+1}\right| +C = \ln |u-1| -\ln |u+1| +C = -2\tanh^{-1} (u) +C$$

Last edited: Jan 14, 2008
14. Jan 15, 2008

fermio

Just type 1/(x sqrt(4x+1)) into http://integrals.wolfram.com/index.jsp and answer will be
$$\int \frac{dx}{x\sqrt{4x+1}}=\ln(\sqrt{4x+1}-1)-\ln(\sqrt{4x+1}+1)+C$$
And hyperbolic arctangent http://mathworld.wolfram.com/InverseHyperbolicTangent.html is
arctanh(z)=0.5[ln(1+z)-ln(1-z)] and (arctanh(z))'=1/(1-z^2)
So this guy looks like integrate correctly, but then it means that wolfram integrator integrate wrong? Becouse according to this guy final answer will be
$$-2\int \frac{du}{1-u^2} =\ln(1-u)-\ln(1+u)+C=\ln(1-\sqrt{4x+1})-\ln(1+\sqrt{4x+1})$$
so answers are different, computer vs human, who lieing?
BTW wolfram integrator showing that 1/(1-x^2) is
$$\int\frac{dx}{1-x^2}=\frac{1}{2}(\ln(x+1)-\ln(x-1))$$
Looks like I found bug in wolfram integrator, becouse everywhere on internet $$(\tanh^{-1} (x))'=\frac{1}{1-x^2}$$

And in this lecture http://www.midnighttutor.com/int_x^2sinxdx.html he missed 2 in final answer.

Last edited: Jan 15, 2008
15. Jan 15, 2008

Kurdt

Staff Emeritus
Sorry yes in post 11 I gave the identity for arccoth (x) by mistake. I think That wolfram may be taking the absolute value in which case it doesn't matter which way round you write them but you have to be aware that arctanh has a particular domain. So in that respect perhaps its not so good.

Last edited: Jan 15, 2008
16. Jan 19, 2008