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Calculus Help for a Friend

Blogs > Nick_54
Post a Reply
Nick_54
Profile Blog Joined November 2007
United States2230 Posts
September 20 2008 06:26 GMT
#1
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.

Day[9]
Profile Blog Joined April 2003
United States7366 Posts
September 20 2008 06:28 GMT
#2
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. www.day9.tv
Nick_54
Profile Blog Joined November 2007
United States2230 Posts
Last Edited: 2008-09-20 06:41:12
September 20 2008 06:32 GMT
#3
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.
il0seonpurpose
Profile Blog Joined January 2007
Korea (South)5638 Posts
September 20 2008 06:43 GMT
#4
dude watch the savior games!!
Nick_54
Profile Blog Joined November 2007
United States2230 Posts
September 20 2008 06:46 GMT
#5
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.
Saracen
Profile Blog Joined December 2007
United States5139 Posts
Last Edited: 2008-09-20 07:37:59
September 20 2008 07:26 GMT
#6
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 brain
JacobDaKung
Profile Blog Joined May 2006
Sweden132 Posts
Last Edited: 2008-09-20 08:23:02
September 20 2008 08:22 GMT
#7
answer is ((x^2)^1.5 - (.5x^2)^1.5)/1.5
Storchen
Profile Joined September 2006
Sweden4385 Posts
September 20 2008 09:12 GMT
#8
Your answer is 5.
5
Nytefish
Profile Blog Joined December 2007
United Kingdom4282 Posts
Last Edited: 2008-09-20 14:37:45
September 20 2008 09:14 GMT
#9
eheh nvm didn't read question right
No I'm never serious.
ZpuX
Profile Blog Joined December 2002
Sweden1230 Posts
Last Edited: 2008-09-21 18:30:00
September 20 2008 12:13 GMT
#10
...
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.
Really, play for fun!
Klockan3
Profile Blog Joined July 2007
Sweden2866 Posts
September 20 2008 14:11 GMT
#11
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
Cambium
Profile Blog Joined June 2004
United States16368 Posts
September 20 2008 14:39 GMT
#12
42
When you want something, all the universe conspires in helping you to achieve it.
ZpuX
Profile Blog Joined December 2002
Sweden1230 Posts
September 20 2008 15:13 GMT
#13
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)....
Really, play for fun!
Klockan3
Profile Blog Joined July 2007
Sweden2866 Posts
September 21 2008 11:04 GMT
#14
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.
ZpuX
Profile Blog Joined December 2002
Sweden1230 Posts
September 21 2008 11:38 GMT
#15
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...
Really, play for fun!
Klockan3
Profile Blog Joined July 2007
Sweden2866 Posts
Last Edited: 2008-09-21 12:40:20
September 21 2008 12:36 GMT
#16
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.
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