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Nick_54
United States2230 Posts
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 States7366 Posts
all you have to do is use the fundamental theorem of calculus to solve it too!!!! | ||
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Nick_54
United States2230 Posts
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. | ||
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il0seonpurpose
Korea (South)5638 Posts
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Nick_54
United States2230 Posts
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 States5139 Posts
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 | ||
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JacobDaKung
Sweden132 Posts
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Storchen
Sweden4385 Posts
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Nytefish
United Kingdom4282 Posts
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ZpuX
Sweden1230 Posts
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. | ||
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Klockan3
Sweden2866 Posts
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
United States16368 Posts
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ZpuX
Sweden1230 Posts
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
Sweden2866 Posts
On September 21 2008 00:13 ZpuX wrote: 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
Sweden1230 Posts
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Klockan3
Sweden2866 Posts
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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