# Integral Problems

• Feb 13th 2007, 06:27 PM
coolio
Integral Problems
uo
• Feb 13th 2007, 06:40 PM
ThePerfectHacker
I am not going to do the definite integral, rather I will find the anti-derivative and leave you to evaluate them at the endpoints.
• Feb 13th 2007, 07:35 PM
ThePerfectHacker
Next one
• Feb 14th 2007, 08:28 AM
ticbol
Quote:

Originally Posted by ThePerfectHacker
I am not going to do the definite integral, rather I will find the anti-derivative and leave you to evaluate them at the endpoints.

Can explain, perfecthacker, in your math ways, why if u = x, then du = 1 ?
Never have I encountered that before in my math ways?

Do you have your own math ways?
• Feb 14th 2007, 08:46 AM
ticbol
Quote:

Originally Posted by ThePerfectHacker
I am not going to do the definite integral, rather I will find the anti-derivative and leave you to evaluate them at the endpoints.

Again, if u' = 1
Then u = x --------------???

Also, if v = arcsin(x)
Then v' = 1 / sqrt(1 -x^2) ------------???

then INT. [u]dv
= INT. [x / sqrt(1-x^2)]dx ---------where did you get dx?
• Feb 14th 2007, 11:20 AM
ThePerfectHacker
Quote:

Originally Posted by ticbol
Can explain, perfecthacker, in your math ways, why if u = x, then du = 1 ?
Never have I encountered that before in my math ways?

Do you have your own math ways?

In my other post in geometry, I talk about "rigorous" and how important it is to mathemations.

Now, the concepts of differenencials, splitting the dy and the dx as if they are fractions.

Like for example,
dy/dx=y
Then,
(1/y)dy=dx
Is not really acceptable among mathemations because they are not fractions. Thus, I use a more formal approach which does not involve splitting the denominator of the differencial.

I wrote a Calculus thread over here.
http://www.mathhelpforum.com/math-he...utorial-2.html

(The math program that is used to generate math code is math working properly now, thus you might not see everything).

In that thread I explain how I do integration via the substitution rule my way.

~~
And yes, that is my own techinque.