Introduction to Calculus Tutorial - Page 3

^_^Engineer_Adam^_^ · #31 $Old$ May 28th, 2007, 06:36 AM

Quote:

Originally Posted by ThePerfectHacker $View Post$

Find the integrals.

4*) $\int e^{\sqrt{x}}$ It has a star what do you think!?

10) $\int \frac{1}{1-e^{2x}}dx$ Hard

hi
i cant seem to solv this two problems can u please provide solution to these?

arrgh

ThePerfectHacker · #32 $Old$ May 28th, 2007, 09:32 AM

Quote:

Originally Posted by ^_^Engineer_Adam^_^ $View Post$

hi
i cant seem to solv this two problems can u please provide solution to these?

arrgh

$\int e^{\sqrt{x}} dx$
I will make the substitution $u=\sqrt{x}$ . But if I do that, I always need to know what its derivative is, $u' = \frac{1}{2\sqrt{x}}$ .

If you look into the integral this factor does not appear. So what do we do? We make it appear. Multiply the numerator and denominator by this expression:

$\int e^{\sqrt{x}}\cdot 2\sqrt{x} \cdot \frac{1}{2\sqrt{x}} dx$
Now if we use $u=\sqrt{x}$ then, $u'=\frac{1}{\sqrt{x}}$ which is good because it appears as a factor and $2\sqrt{x} = 2u$ .
Make the substitution,
$\int e^u \cdot 2u \cdot u' dx = \int 2u e^u du$ .
You can now do this integral by parts.

^_^Engineer_Adam^_^ · #33 $Old$ May 30th, 2007, 04:47 PM

i get it now
$\int \frac{1}{1-e^{2x}}dx$

$\int \frac{1 - e^{2x}+e^{2x}}{1-e^{2x}}dx$

$\int \frac{1-e^{2x}}{1-e^{2x}}dx + \frac{e^{2x}}{1-e^{2x}}dx$

$x + \int \frac{e^{2x}}{u} \frac{du}{-2e^{2x}}$

$x - \frac{1}{2}\ln(1-e^{2x}) +C$

but the integrator says

$x - \frac{1}{2}\ln(e^{2x}-1)$
hmm

ThePerfectHacker · #34 $Old$ May 30th, 2007, 04:58 PM

Quote:

Originally Posted by ^_^Engineer_Adam^_^ $View Post$

i get it now
$\int \frac{1}{1-e^{2x}}dx$

$\int \frac{1 - e^{2x}+e^{2x}}{1-e^{2x}}dx$

$\int \frac{1-e^{2x}}{1-e^{2x}}dx + \frac{e^{2x}}{1-e^{2x}}dx$

$x + \int \frac{e^{2x}}{u} \frac{du}{-2e^{2x}}$

$x - \frac{1}{2}\ln(1-e^{2x}) +C$

but the integrator says

$x - \frac{1}{2}\ln(e^{2x}-1)$
hmm

Because you need to use $| \, |$ .
In that case,
$\ln |1-e^{2x}| = \ln |e^{2x}-1|$
Because,
$|a-b|=|b-a|$

tukeywilliams · #35 $Old$ June 30th, 2007, 01:06 AM

just wanted to point out: isn't a sequence a function defined as $f: \mathbb{Z^{+}} \rightarrow A$ which means a sequence in the set $A$ ? So the codomain doesn't have to be $\mathbb{R}$ but can be an arbitrary set $A$ ? Actually $\mathbb{N}$ and $\mathbb{Z^{+}}$ are equivalent right? And a sequence is null when $\lim_{n \rightarrow \infty} f(n) = 0$ when $\forall \epsilon \in \mathbb{R^{+}}, \exists N \in \mathbb{Z^{+}}, \forall n \in \mathbb{Z^{+}} (n \geq N \Rightarrow |f(n)| < \epsilon)$ . Also, I like to think of functions as follows: pretend you have a box subdivided into smaller boxes. These smaller boxes represent the elements of the codomain, and the points in the boxes represent the elements of the domain. So a function orders the points in the smaller boxes.

ThePerfectHacker · #36 $Old$ June 30th, 2007, 07:58 PM

Quote:

Originally Posted by tukeywilliams $View Post$

just wanted to point out: isn't a sequence a function defined as $f: \mathbb{Z^{+}} \rightarrow A$ which means a sequence in the set $A$ ? So the codomain doesn't have to be $\mathbb{R}$ but can be an arbitrary set $A$ ? Actually $\mathbb{N}$ and $\mathbb{Z^{+}}$ are equivalent right? And a sequence is null when $\lim_{n \rightarrow \infty} f(n) = 0$ when $\forall \epsilon \in \mathbb{R^{+}}, \exists N \in \mathbb{Z^{+}}, \forall n \in \mathbb{Z^{+}} (n \geq N \Rightarrow |f(n)| < \epsilon)$ . Also, I like to think of functions as follows: pretend you have a box subdivided into smaller boxes. These smaller boxes represent the elements of the codomain, and the points in the boxes represent the elements of the domain. So a function orders the points in the smaller boxes.

I was talking about a "real sequence" in that case $f:\mathbb{N}\to \mathbb{R}$ .

colby2152 · #37 $Old$ January 4th, 2008, 01:45 PM

TPH, your notes are extremely thorough and detailed, and are an excellent source for anyone studying Calculus. I hope you don't mind that I added some of my "simpler" Calculus notes to this forum. Like yours, my notes are not meant for students taking Advanced Calculus courses such as Real Analysis. I hope I am not stepping on your toes!

Jhevon · #38 $Old$ January 4th, 2008, 01:49 PM

Quote:

Originally Posted by colby2152 $View Post$

TPH, your notes are extremely thorough and detailed, and are an excellent source for anyone studying Calculus. I hope you don't mind that I added some of my "simpler" Calculus notes to this forum. Like yours, my notes are not meant for students taking Advanced Calculus courses such as Real Analysis. I hope I am not stepping on your toes!

that would be all good and well i believe. in fact, we encourage users to do things like that. that's why we have a mathwiki page (but it's not fully functional yet). as long as you're not "re-inventing the wheel" (that is, going over exactly the same stuff TPH covered, unless of course, you feel he left something out) i don't see why you can't post notes. you may want to do it in a different thread though, this is TPH's thread.

mustkara · #39 $Old$ February 7th, 2008, 04:13 AM

I believe the textbook

Introduction to Stochastic Calculus with Applications, 1st Edition

will be useful for you to study Mathematics

Good Luck and wish help for you.

hehe ^_^

scorpion007 · #40 $Old$ May 15th, 2009, 03:24 AM

Just some spelling corrections: Differenciable, Differencial, Differenciation should all be Differentiable, Differential, Differentiation respectively. Basically, the C's should be T's.

Moo · #41 $Old$ May 15th, 2009, 12:48 PM

Quote:

Originally Posted by scorpion007 $View Post$

Just some spelling corrections: Differenciable, Differencial, Differenciation should all be Differentiable, Differential, Differentiation respectively. Basically, the C's should be T's.

Oh, that's just TPH, you shouldn't bother that much for it

Another example that comes in mind is that he once said "diagnol" for "diagonal". It's common, but it doesn't alter the quality