 Nick_54 United States. September 20 2008 15:26. Posts 978 | Profile Blog |
So my one friends is taking a caluculus class that is beyone my level. I took basic calc at a junior college, but now he is in the second class. So I am asking TL for help. After reading that IQ thread a while ago it should be easy. Let me see if I remember the notation right.
Its an integral the upper limit is x^2 and the lower limit is (x^2)/2. The inside of the integral is ln t^.5 dt.
Sorry if that is hard to read I don't know of any converter. Thanks again for your help and I will eventually post one of these where I am just not begging for help.
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 Day[9] United States. September 20 2008 15:28. Posts 6304 | Profile Blog |
this is a cute problem
all you have to do is use the fundamental theorem of calculus to solve it too!!!! |
| | Whenever I encounter some little hitch, or some of my orbs get out of orbit, nothing pleases me so much as to make the crooked straight and crush down uneven places |
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| Nick_54 United States. September 20 2008 15:32. Posts 978 | Profile Blog |
On September 20 2008 15:28 Day[9] wrote:
this is a cute problem
all you have to do is use the fundamental theorem of calculus to solve it too!!!!
Is it 0? Idk haven't taken it in like 2 years so I don't remember anything lol.
Edit: Day if Savior wins this match you have to help me lol.Last edit: 2008-09-20 15:41:12 |
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 il0seonpurpose Korea (South). September 20 2008 15:43. Posts 5213 | Profile Blog |
| dude watch the savior games!! |
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| Nick_54 United States. September 20 2008 15:46. Posts 978 | Profile Blog |
On September 20 2008 15:43 il0seonpurpose wrote:
dude watch the savior games!!
I am watching it, but my comp can't handle the stream. |
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 Saracen United States. September 20 2008 16:26. Posts 2310 | Profile Blog |
On September 20 2008 15:43 il0seonpurpose wrote:
dude watch the savior games!!
+ Show Spoiler +
don't
it will make you cry
EDIT: i feel so dumb right now ... can't do math anymore
.5x^2(lnx^2-1) - .25x^2(ln(x^2/2)-1)
simplify gogo
god that better be right
2:40 AM and summer vacation do absolute shit for your brainLast edit: 2008-09-20 16:37:59 |
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 JacobDaKung Sweden. September 20 2008 17:22. Posts 44 | Profile Blog |
| answer is ((x^2)^1.5 - (.5x^2)^1.5)/1.5 Last edit: 2008-09-20 17:23:02 |
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 Storchen Sweden. September 20 2008 18:12. Posts 4211 | Profile | |
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 Nytefish United Kingdom. September 20 2008 18:14. Posts 1846 | Profile Blog |
| eheh nvm didn't read question right Last edit: 2008-09-20 23:37:45 |
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| ZpuX Sweden. September 20 2008 21:13. Posts 977 | Profile Blog |
...
ln(X^0.5) = (1/2)ln(X), the prime function of ln(x) is xln(x) - x if I remember correctly
so (1/2)*[xln(x) - x)] with the limit x^2/2 --> x^2, simplified should be:
((X^2)/4) * (1+ ln(X^2) + ln(2)) final answer.Last edit: 2008-09-22 03:30:00 |
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 Klockan3 Sweden. September 20 2008 23:11. Posts 1882 | Profile Blog |
This should be correct unlike the other answers:
Substitute sqrt(t)=M
dM*2M=dt
int(ln(sqrt(t))dt=int(ln(M)*2MdM)
primitive function to that is:ln(M)*M^2-(M^2)/2
Boundary changes to due integration leads to:
ans=ln(x)*x^2-(x^2)/2-ln(x/2)*(x/2)^2+(x^2)/8=0.75*ln(2x)x^2-(x^2)3/8
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 Cambium Canada. September 20 2008 23:39. Posts 9999 | Profile Blog | |
| | When you want something, all the universe conspires in helping you to achieve it. |
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| ZpuX Sweden. September 21 2008 00:13. Posts 977 | Profile Blog |
On September 20 2008 23:11 Klockan3 wrote:
This should be correct unlike the other answers:
Substitute sqrt(t)=M
dM*2M=dt
int(ln(sqrt(t))dt=int(ln(M)*2MdM)
primitive function to that is:ln(M)*M^2-(M^2)/2
Boundary changes to due integration leads to:
ans=ln(x)*x^2-(x^2)/2-ln(x/2)*(x/2)^2+(x^2)/8=0.75*ln(2x)x^2-(x^2)3/8
but your boundary changes are wrong...
we have t = (x^2)/2, and M = sqrt(t) ---> M = sqrt(x^2/2) = x/sqrt(2).... |
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| Klockan3 Sweden. September 21 2008 20:04. Posts 1882 | Profile Blog |
On September 21 2008 00:13 ZpuX wrote:
Show nested quote +On September 20 2008 23:11 Klockan3 wrote:
This should be correct unlike the other answers:
Substitute sqrt(t)=M
dM*2M=dt
int(ln(sqrt(t))dt=int(ln(M)*2MdM)
primitive function to that is:ln(M)*M^2-(M^2)/2
Boundary changes to due integration leads to:
ans=ln(x)*x^2-(x^2)/2-ln(x/2)*(x/2)^2+(x^2)/8=0.75*ln(2x)x^2-(x^2)3/8
but your boundary changes are wrong... we have t = (x^2)/2, and M = sqrt(t) ---> M = sqrt(x^2/2) = x/sqrt(2)....
Yeah, they are 
But its just in the last steps. |
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| ZpuX Sweden. September 21 2008 20:38. Posts 977 | Profile Blog |
| well dont try to sound so superior, and recheck your answer with the others, and you'll see if you actually were the "only" right... |
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| Klockan3 Sweden. September 21 2008 21:36. Posts 1882 | Profile Blog |
On September 21 2008 20:38 ZpuX wrote:
well dont try to sound so superior, and recheck your answer with the others, and you'll see if you actually were the "only" right...
But your answer is wrong on so many places that I did not read through the post throughout, but it seems like all the errors were made when you tried to calculate it and not earlier. Also you should try to avoid confusing notations such as x^2/2, it was a part why I skimmed your post since I thought you meant something different than (x^2)/2. I use it now in this to make it easier to follow from yours but you should make yourself clear.
(1/2)*[xln(x) - x)] with the limit x^2/2 --> x^2, simplified should be:
(1/2)*((x^2)*ln(X^3) + (X^2)/2 - ((X^2)/2)*ln(2)).
Is not correct at all. It should be:
(x^2ln(x^2)-x^2)/2-(x^2/2ln(x^2/2)-x^2/2)/2=x^2/2(ln(x^2)-ln(x^2/2)/2)-x^2/2+x^2/4
=x^2/2*ln(2x^2)-x^2/4=ln(x*sqrt2)x^2-1/4*x^2
Or if we write it in more equal terms as yours:
(1/2)*((x^2)*ln(X^2) - (X^2)/4 + ((X^2)/2)*ln(2)).
Also this is the same as you would get if you used the correct boundary change in my own. Last edit: 2008-09-21 21:40:20 |
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