# Introduction to Calculus Tutorial

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• May 28th 2007, 04:36 AM
Quote:

Originally Posted by ThePerfectHacker

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? :D
arrgh
• May 28th 2007, 07:32 AM
ThePerfectHacker
Quote:

hi
i cant seem to solv this two problems can u please provide solution to these? :D
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.
• May 30th 2007, 02:47 PM
solution for number 10
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
• May 30th 2007, 02:58 PM
ThePerfectHacker
Quote:

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,
$|a-b|=|b-a|$
• Jun 29th 2007, 11:06 PM
tukeywilliams
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.
• Jun 30th 2007, 05:58 PM
ThePerfectHacker
Quote:

Originally Posted by tukeywilliams
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 $f:\mathbb{N}\to \mathbb{R}$.
• Jan 4th 2008, 11:45 AM
colby2152
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! (Handshake)
• Jan 4th 2008, 11:49 AM
Jhevon
Quote:

Originally Posted by colby2152
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! (Handshake)

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.
• Feb 7th 2008, 02:13 AM
mustkara
Good Reference Textbook
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 ^_^
• May 15th 2009, 01:24 AM
scorpion007
Just some spelling corrections: Differenciable, Differencial, Differenciation should all be Differentiable, Differential, Differentiation respectively. Basically, the C's should be T's.
• May 15th 2009, 10:48 AM
Moo
Quote:

Originally Posted by scorpion007
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 (Rofl)

Another example that comes in mind is that he once said "diagnol" for "diagonal". It's common, but it doesn't alter the quality :D
• Dec 10th 2009, 08:58 AM
rainer
On the off chance that TPH or some moderator is still attending responses to this thread--

Regarding elliptical integrals TPH dixit: "Now the problem is that this integral cannot be expressed in elementary functions (the standard functions) like I mentioned in the previous lecture that that sometimes happens. This problem was studied primarily by Abel and lead to something call Elliptic Functions. These are new types of functions that integrate this. I myself do not know what courses actually teach these functions, not because they are difficult but because it is unnecessary."

And elsewhere, regarding the integration of radicals (particularly the arclength formula):

"We did not discuss how to integrate that, it can be done. The types of functions used are hyperbolic functions, they are related to the exponental e^x function. But that is too advacned for us, and all we did was set up the integral."

I would like to look into these related matters a little more deeply. Can TPH or anyone recommend a text please?

Oh yes, and many thanks indeed to the author of this awesome thread.
• Jan 8th 2010, 08:07 PM
TWiX
Quote:

Originally Posted by ThePerfectHacker
11*)Show that if $\sum_{n=0}\frac{kn}{n^2+1}$ converges only when $k=0$.

1-if k < 0 then the series diverges by the L.C.T with -1/n
2-if k > 0 then the series diverges by the limit comparison test with 1/n

If k=0 the series converges since $\sum_{}0$ converges.

but 1 is wrong
since -1/n is not positive

• Jan 9th 2010, 03:36 PM
Jhevon
Quote:

Originally Posted by TWiX
1-if k < 0 then the series diverges by the L.C.T with -1/n
2-if k > 0 then the series diverges by the limit comparison test with 1/n

If k=0 the series converges since $\sum_{}0$ converges.

but 1 is wrong
since -1/n is not positive

you could write $k = -m$ where $m>0$ and consider the series $\sum_{n=0}^\infty \frac {mn}{n^2 + 1}$. By 2 it diverges. Hence the series $\sum_{n=0}^\infty \frac {kn}{n^2 + 1} = - \sum_{n=0}^\infty \frac {mn}{n^2 + 1}$ diverges.
• Jan 10th 2010, 12:20 PM
TWiX
Quote:

Originally Posted by ThePerfectHacker
$\int e^{\sqrt{x}} dx$
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.

Sorry, But this is wrong
You should prove that x dont equal to zero first
you cant devide by a variable If you dont prove it dont equal to 0
Substitute $u=\sqrt{x}$
then
$u^2 = x$
$dx=2udu$
$\int e^{\sqrt{x}} dx = 2 \int u e^{u} du$
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