[maths] Re: integrals; calculus
- From: "Ned Granic" <ngranic@xxxxxxx>
- To: <maths@xxxxxxxxxxxxx>
- Date: Tue, 26 Dec 2006 13:05:17 -0700
Hi Nelson,
and merry xmas to you as well! All the best in the new year.
I cannot brag about any success in my calc class since I got a C; I wish I got
an F so I could correct it, and it makes me really mad.
However, it's over.
The only remaining formula for sums of powers of positive integers I have is
when k = 2, and it is:
$_(i=1)~n i^2 = {[n(n+1)(2n+1)] / 6}.
There might be more of them in one of the apendixes, but I have to check.
I used that example as an illustration how to solve the equation:
&_-1~5 (1+3x) dx.
In (1+3x) it's the number 1 that confuses me, and I don't know how to treat it.
Every way I tried, it gave me a wrong solution.
When using the fundamental theorem of calculus, I get the correct result.
But anyway, the textbook I have is very interesting, and I intend to study it
further on my own if there will be some time available for it. So, you might
find me again on this list asking a question here and there even though my
required math studies are over.
I thank you again for all your support!
Cheers!
Ned
----- Original Message -----
From: Nelson Blachman
To: maths@xxxxxxxxxxxxx
Sent: Monday, December 25, 2006 11:02 AM
Subject: [maths] Re: integrals; calculus
Ned,
I've finally found enough time to study your marvelous evaluation of the
integral of (x^3 - 6x) dx from 0
to 3, which seems quite correct because you were able to do something I
couldn't.
You made use of the fact that the summation of i^3 from i=0 to i=n is
[n(n+1)]^2, a fact that I'd never learned but which you were somehow able
to discover and put to good use here.
Do you have formulas for the summation of i^k from i=1 to i=n for other
values of k?
I see you used n(n+1)/2 for k=1, a fact of which I've long been quite
aware, since I can derive it veary easily as n times the mean value of i.
Merry Xmas,
Nelson
----- Original Message -----
From: Ned Granic
To: maths@xxxxxxxxxxxxx
Sent: Thursday, December 14, 2006 2:16 PM
Subject: [maths] Re: integrals; calculus
Hi Nelson and thank you.
Well, my calculus class is over.
I'll know how I did in a week or so.
I thank you again for all your help which was sometimes crucial.
As far as my last two questions go, here is one example of a problem that
relates to my 1st question.
Let me first explain my notation since writing in words would be extremely
extensive and confusing, and there is no standard one yet, but I guess my
invention works for me.
Oops! I use Braill Lite 40 for math that has 8-dot mode, and here on the
Windows box, it's gonna be tough. Reinventing the wheal again.
Well: try this
& ... the integral sign
_ ... indicating that the following should be underneath the sign it follows
~ ... sign above; opposite of the _
; -- the sub sign; opposite of ^ (to the power of)
-> ... right-arrow; approaching sign
I ... positive infinity
$ ... sigma; summation notation
For instance, the formula for the definition of the integral says:
&_a~b f(x) dx = lim_(n->I) $_(i=1)~n f(x;i) delta x.
and it's read as:
Integral from a to b of f of x dx equals the limit as n approaches infinity
of the summation from i equals 1 to n of f of x sub i times delta x.
That's enough for the integrals. So:
Evaluate:
&_0~3 (x^3 - 6x) dx.
delta x = (b-a)/n
= 3/n
x;0 = 0, x;1 = 3/n, x;2 = 6/n, ... x;i = (3i)/n.
&_0~3 (x^3 - 6x) dx
= lim_(n->I) $_(i=1)~n f[(3i)/n] 3/n
= lim_(n->I) 3/n$_(i=1)~n [(3i/n)^3 - 6(3i/n)]
= lim_(n->I) 3/n$_(i=1)~n [27/n^3 i^3 - 18/n i]
= lim_(n->I) [81/n^4 * $_(i=1)~n i^3 - 54/n^2 * $_(i=1)~n i]
= lim_(n->I) {81/n^4 * [n(n+1)/2]^2 - 54/n^2 * [n(n+1)/2]}
= lim_(n->I) {81/4[(1+1)/n]^2 - 27[(1+1)/n]}
= 81/4-27
= -27/4
= -6.75.
Hope it's not that confusing, particularly if you have a Braille display,
listening to this is just not the way to go.
Cheers!
Ned
----- Original Message -----
From: Nelson Blachman
To: maths@xxxxxxxxxxxxx
Sent: Thursday, December 14, 2006 2:19 AM
Subject: [maths] Re: integrals; calculus
Ned,
I get 42 for the integral of (1+3x)dx from -1 to 5 by using
"antiderivatives." What result do you get via the defining limit and how do
you get it?
For integral from 1 to 64 (1+cube_root x) / sqrt x dx
= integral of [x^(-1/2) + x^(-1/6)] dx =
= [2x^(1/2) + (6/5)x^(6/5)] for x=64 minus for x=1= 14 + 37.2 = 51.2
What do you get, and how?
I haven't given any thought yet to how to use the defining limit.
--Nelson
----- Original Message -----
From: Ned Granic
To: maths@xxxxxxxxxxxxx
Sent: Wednesday, December 13, 2006 11:23 AM
Subject: [maths] integrals; calculus
Hi all,
A couple of very quick questions:
1. Compute the following using the equation as limits of sums from the
definition of the integral.
That equation is:
integral from a to b of f(x) dx
= limit as n approaches infinity of summation (Sigma) from i=1 to n of
f(x sub i) delta x.
The problem itself is:
integral from -1 to 5 of (1+3x) dx.
When I use the FTC2, I get the correct answer 42, but not this way.
2. Use FTC2 to evaluate:
integral from 1 to 64 (1+cube_root x) / sqrt x dx.
I get 32, but the right answer is 256/5.
Many thanks in advance!
Ned
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