Mathematics Corner :)

math maTH MATHHHHHHHHHHH ... horrible memories when it came to studying but thak god i MADE IT.. anyways im gonna be taking math next semester to get it outta the WAY .. so yall will expect many Q's to come .. n i hope yall will help me .. so i might be too much lol but i do need help in this area.. it normally ends me up with a headache.. i cant understand how ppl enjoy it.anywayz thank GOD we have a math corner YAY .. thanks to who ever came up with SUCH A WONDERFULLLLLLL idea.. :))))
 
Whatsthepoint,if you're not in UNI you should be playing with pencil crayons. :D anyway which grade does calcus really start?
 
eh!!!!!!!! stumbled in here expecting to see something about numbers, seem to have inadvertantly witnessed some sort of alien language. my head hurts im off to the non genius end of the forum:-[
 
Thats the funniest thing I heard on this forum so far:D:D:D:D:D:D:D
 
Yanal said:
Whatsthepoint,if you're not in UNI you should be playing with pencil crayons. :D anyway which grade does calcus really start?
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It's quite funny, I only started calculus in year 10..was about 14/15ish. When are you expected to start calculus in the states do you know?
 
Alpha Dude said:
Although I've forgotten most of it now, I use to love calculus.

You're a brother - you're not allowed access to the sisters' section. :-[

Muhahahaha.
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That's why it's a place where you feel at home and I feel out of place.

Can someone do this:

Show that the integral of 1/(1+x^2) dx between the limits of 1 and 0 = pi/4 [Let x = tan u]

Integration by subtitution. I hate that part of it.
 
Sure....though it's gonna be like impossible to understand in this format...

if x = tan u, dx/du = sec^2 u --> dx = sec^2 u du

and 1 + x^2 = 1+tan^2 u = sec^2 u

So, the integral of 1/(1+x^2) dx
= integral of (1/sec^2 u) dx
= integral of (1/sec^2 u)*sec^2 u du
= integral of 1 du =

If the limits of the original integral (in x) were 0 and 1, and x = tanu --> u=invtan x.
The limits of the new integral are invtan(0) = 0, and invtan(1) = pi/4.

Putting these limits for gives [(pi/4) - (0)] = pi/4.

Voila!
 
If the limits of the original integral (in x) were 0 and 1, and x = tanu --> u=invtan x.
The limits of the new integral are invtan(0) = 0, and invtan(1) = pi/4.
Click to expand...

Ah, I see... That's where I messed up.
 
Dude do some math understandable for all suitable ages or put the title as 18+ only.
 
I seee....the limits are definitely the most annoying part of integration by substitution, lol...
 
Okay... Here's another one that I got stuck on. And I also tried it several times in several ways but never got to the answer at the back of the book:

The integral of x/(2x+5)^1/2 between the limits of 10 and 2
Let u^2 = (2x+5)

Again, integration by substitution. And the answer at the back of the book = 34/3.
 
Okey dokey...

Firstly, u^2 = (2x+5) so u =sqrt(2x+5).
Therefore, du/dx = (2x+5)^-0.5 = 1/sqrt(2x+5) --> dx = sqrt(2x+5) du = u du

If the limits of the original integral, in x, are 10 and 2, the integral in u has the limits sqrt(2(10)+5) = 5, and sqrt(2(2)+5) = 3.

Also, we need to rearrange u^2 = (2x+5) to get x = 0.5(u^2-5), which we also substitute into the integral.

The integral of x/(2x+5)^1/2 dx therefore simplifies to the integral of (0.5(u^2-5))/u * u du, which is [0.5((u^3)/3 - 5u)]. I'm skipping a few steps, but evaluate this between the limits 5 and 3, and you'll get 34/3...
 
I had to do that on paper before it was clear. But thanks anyway.

Are there any particular times when you do du/dx rather than dx/du when substituting? Does it make a difference?
 
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!:exhausted
 
~Raynn~ said:
Okey dokey...

Firstly, u^2 = (2x+5) so u =sqrt(2x+5).
Therefore, du/dx = (2x+5)^-0.5 = 1/sqrt(2x+5) --> dx = sqrt(2x+5) du = u du

If the limits of the original integral, in x, are 10 and 2, the integral in u has the limits sqrt(2(10)+5) = 5, and sqrt(2(2)+5) = 3.

Also, we need to rearrange u^2 = (2x+5) to get x = 0.5(u^2-5), which we also substitute into the integral.

The integral of x/(2x+5)^1/2 dx therefore simplifies to the integral of (0.5(u^2-5))/u * u du, which is [0.5((u^3)/3 - 5u)]. I'm skipping a few steps, but evaluate this between the limits 5 and 3, and you'll get 34/3...
Click to expand...
Hey lady, are you a mathematician?
 
The Ruler said:
I had to do that on paper before it was clear. But thanks anyway.

Are there any particular times when you do du/dx rather than dx/du when substituting? Does it make a difference?
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I'm sorry, yeah, I squeezed several steps into one line...and nope, it doesn't make any difference (and anyway, du/dx is just 1/(dx/du))....but if you're given an expression for the substitution in the form x = [something]u (like x = tanu) then it's more natural to keep it that way, and find dx/du.....if instead, you have u = (some expression in x), you'd just find du/dx...

Whatsthepoint said:
Hey lady, are you a mathematician?
Click to expand...
:giggling: Welllllll, I do A level maths and further maths...does that count?? :D
 
*Whoosh!*

^That is the sound of Rayyn's mathematics going straight over my head.
 
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