View Full Version : Help me with this math problem.
My Opinion Matters
6/14/2011, 04:59 PM
(1/2,0)∫ 5 sin^7 2∏x dx
Show your work; but remember, this board has a 25000 character limit for each post.
Aaaaand go! :texan:
NormanPride
6/14/2011, 05:01 PM
http://homepage.eircom.net/~miscellaneous/images/animath.jpg
yermom
6/14/2011, 05:03 PM
what is (1/2,0)?
and should it be 5*sin^7(2*pi*x) dx?
My Opinion Matters
6/14/2011, 05:04 PM
what is (1/2,0)?
should it be 5*sin^7(2*pi*x) dx?
Those are the limits of integration, you silly man.
OUMallen
6/14/2011, 05:05 PM
I wish I were your derivative so that I could lie tangent to your curves.
yermom
6/14/2011, 05:06 PM
Those are the limits of integration, you silly man.
.5 to 0?
My Opinion Matters
6/14/2011, 05:06 PM
And yes, 5 times sin raised to the 7th power times (2pi*x), integrated. Quit stalling!
My Opinion Matters
6/14/2011, 05:07 PM
.5 to 0?
Yes.
yermom
6/14/2011, 05:09 PM
i don't know if i care enough to go this far back in time :D
soonerchk
6/14/2011, 05:10 PM
3. The answer is always 3.
My Opinion Matters
6/14/2011, 05:18 PM
3. The answer is always 3.
I don't think this is right.
soonerchk
6/14/2011, 05:21 PM
I don't think this is right.
Then it's 7. Those are the only options.
Fraggle145
6/14/2011, 05:29 PM
http://homepage.eircom.net/~miscellaneous/images/animath.jpg
This doesnt work. -7 doesnt equal 19. Unless there are subtle differences in the x's that I'm not catching.
Thaumaturge
6/14/2011, 05:46 PM
That's a pretty standard type of integral. Why would you come to these rubes?
GKeeper316
6/14/2011, 05:52 PM
the answer is 42. duh.
Howzit
6/14/2011, 05:52 PM
3. The answer is always 3.
B got me through Physics 2 in college.
MR2-Sooner86
6/14/2011, 06:03 PM
I'll get you started:
∫sin^5(x)dx
∫sin(x)sin^2(x)sin^2(x)dx
∫sin(x) (1-cos^2(x)) (1-cos^2(x)) dx
∫sin(x) (1 - 2cos^2(x) + cos^4(x)) dx
∫(1 - 2(cos(x))^2 + (cos(x))^4) sin(x)dx
Let u = cos(x). Then du = -sin(x)dx. So, we have:
∫(1 - 2u^2 + u^4) * -du
-∫(1 - 2u^2 + u^4)du
-(u - 2(u^3)/3 + (u^5)/5)
I think you can take it from there.
crawfish
6/14/2011, 06:04 PM
Can I use imaginary numbers?
GKeeper316
6/14/2011, 06:07 PM
42
My Opinion Matters
6/14/2011, 06:20 PM
I'll get you started:
∫sin^5(x)dx
∫sin(x)sin^2(x)sin^2(x)dx
∫sin(x) (1-cos^2(x)) (1-cos^2(x)) dx
∫sin(x) (1 - 2cos^2(x) + cos^4(x)) dx
∫(1 - 2(cos(x))^2 + (cos(x))^4) sin(x)dx
Let u = cos(x). Then du = -sin(x)dx. So, we have:
∫(1 - 2u^2 + u^4) * -du
-∫(1 - 2u^2 + u^4)du
-(u - 2(u^3)/3 + (u^5)/5)
I think you can take it from there.
Close, but no.
Partial Qualifier
6/14/2011, 07:10 PM
Given I(n) = [-sin^(n-1)(x) cos x]/n + ((n-1)/n) I(n-2):
Let J(n) = ∫(0 to π/2) sin^n(x) dx = I(π/2) - I(0).
So, we obtain for n > 1:
J(n) = [-sin^(n-1)(x) cos x]/n {for x = 0 to π/2} + ((n-1)/n) J(n-2)
==> J(n) = 0 + ((n-1)/n) J(n-2)
==> J(n) = ((n-1)/n) J(n-2).
----------------------
Now, we use this repeatedly.
J(n) = (n-1)/n J(n-2)
......= ((n-1)/n) * ((n-3)/(n-2)) * J(n-4)
......= ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * J(n-6)
and so on.
----------
If n is odd, we can reduce this all the way down to
J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (2/3) J(1).
However, J(1) = ∫(0 to π/2) sin^1(x) dx = 1.
Hence, J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (2/3).
------------------
If n is even, we can reduce this all the way down to
J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (1/2) J(0).
However, J(0) = ∫(0 to π/2) sin^0(x) dx = π/2.
Hence, J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (1/2) * (π/2).
I hope this helps.
NEEERRRDDDSSS!!!
http://3.bp.blogspot.com/_I2i8xqihwqQ/TDKwKDZiTJI/AAAAAAAAAXU/PBKhmWc2Dqo/s320/revenge-of-the-nerds.jpg
Howzit
6/14/2011, 09:18 PM
Given I(n) = [-sin^(n-1)(x) cos x]/n + ((n-1)/n) I(n-2):
Let J(n) = ∫(0 to π/2) sin^n(x) dx = I(π/2) - I(0).
So, we obtain for n > 1:
J(n) = [-sin^(n-1)(x) cos x]/n {for x = 0 to π/2} + ((n-1)/n) J(n-2)
==> J(n) = 0 + ((n-1)/n) J(n-2)
==> J(n) = ((n-1)/n) J(n-2).
----------------------
Now, we use this repeatedly.
J(n) = (n-1)/n J(n-2)
......= ((n-1)/n) * ((n-3)/(n-2)) * J(n-4)
......= ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * J(n-6)
and so on.
----------
If n is odd, we can reduce this all the way down to
J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (2/3) J(1).
However, J(1) = ∫(0 to π/2) sin^1(x) dx = 1.
Hence, J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (2/3).
------------------
If n is even, we can reduce this all the way down to
J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (1/2) J(0).
However, J(0) = ∫(0 to π/2) sin^0(x) dx = π/2.
Hence, J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (1/2) * (π/2).
I hope this helps.
Beat me to it.
My Opinion Matters
6/14/2011, 09:38 PM
Given I(n) = [-sin^(n-1)(x) cos x]/n + ((n-1)/n) I(n-2):
Let J(n) = ∫(0 to π/2) sin^n(x) dx = I(π/2) - I(0).
So, we obtain for n > 1:
J(n) = [-sin^(n-1)(x) cos x]/n {for x = 0 to π/2} + ((n-1)/n) J(n-2)
==> J(n) = 0 + ((n-1)/n) J(n-2)
==> J(n) = ((n-1)/n) J(n-2).
----------------------
Now, we use this repeatedly.
J(n) = (n-1)/n J(n-2)
......= ((n-1)/n) * ((n-3)/(n-2)) * J(n-4)
......= ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * J(n-6)
and so on.
----------
If n is odd, we can reduce this all the way down to
J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (2/3) J(1).
However, J(1) = ∫(0 to π/2) sin^1(x) dx = 1.
Hence, J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (2/3).
------------------
If n is even, we can reduce this all the way down to
J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (1/2) J(0).
However, J(0) = ∫(0 to π/2) sin^0(x) dx = π/2.
Hence, J(n) = ((n-1)/n) * ((n-3)/(n-2)) * ((n-5)/(n-4)) * ... * (1/2) * (π/2).
I hope this helps.
I am going to have to say this is probably right...because I am not adequately qualified to say that it is not.
yermom
6/14/2011, 09:40 PM
Close, but no.
it was close enough that i remembered how to do it
tommieharris91
6/14/2011, 09:45 PM
This thread needs more politics and mundane ****-flinging.
soonerboomer93
6/14/2011, 09:49 PM
Wtf is Ike when we need a math guy
yermom
6/14/2011, 09:51 PM
This thread needs more politics and mundane ****-flinging.
Leibniz was a little bitch, and couldn't sniff Newton's jock
yermom
6/14/2011, 09:55 PM
Wtf is Ike when we need a math guy
meh. this is Calc II. this is below me, i know it's below Ike :D
i've had literally 7 or 8 math courses since doing this...
Partial Qualifier
6/15/2011, 11:25 AM
Leibniz was a little bitch, and couldn't sniff Newton's jock
....and Einstein wiped his *** with Principia!1!!one!11!!!! ;)
badger
6/15/2011, 12:36 PM
42
Ruuuuuuuuuuuuuuuuuuu!
The answer on defense was always 42 back then :D
stoopified
6/15/2011, 04:53 PM
0.I did it all in my head.
sanantoniosooner
6/15/2011, 05:00 PM
0.I did it all in my head.
dang if this isn't a softball:D
StoopTroup
6/15/2011, 05:06 PM
Anyone have Ginger lee's Digits?
badger
6/15/2011, 05:07 PM
http://static.gamesradar.com/images/mb/GamesRadar/us/Games/P/Professor%20Layton%20and%20the%20Unwound%20Future/Bulk%20Viewer/DS/2010-06-15/NTR_PLaytonUF_06ss07_E3--article_image.jpg
swardboy
6/15/2011, 09:00 PM
E= Mc2
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