x=2t y=3t^2 0<= t <= 1 Evaluate the integral

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x=2t y=3t^2 0<= t <= 1 Evaluate the integral

Mathematics
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\[\int\limits_{}^{}(x-y)ds\]
surface area?
ok

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Other answers:

what is ds?
ds is the line integral which equals sqrt( (dx/dt)^2 + (dy/dt)^2 dt)
ok cool give me just a sec i will scan in something
okay
is that dt outside the sqrt?
no its underneath atleast thats what it seems like in my book go here:http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsPtI.aspx The equation is the 2nd blue one
its outside
okay it just looks like its inside
i had to use integration by parts. it is long problem and maybe i'm going about the long way and there is a shorter way, but I got you started
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don't do it htis way there has to be a shorter way
lol
haha I really appreciate the help, that is exactly where I got to in the problem
I got to \[2\int\limits_{0}^{1}(2t-3t^2)(\sqrt{9t^2+1}\] and after that is where I am getting stuck trying to evaluate the integral
we could try integration by parts but that is messy I have the answer do you want it?
-(1/6)*t*(1+9*t^2)^(3/2)+(1/12)*t*sqrt(1+9*t^2)+(1/36)*arcsinh(3*t)+(4/27)*(1+9*t^2)^(3/2)
yeah that is pretty messy it seems like it should be a lot easier than what we have you know. I got the same thing off of wolframalpha
even the answer with the limits is messy
yeah I know right and this is just a take home quiz idk. Its just that square root that is making messy like that
wait...
omg i got it one sec let me write it neatly
haha okay I sure hope so...my brain is fried lol
ok im posting this but i'm pretty sure i made a mistake somewhere but the problem is easy than we thought
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that is certainly a lot easier than how we were doing it and a much cleaner answer
i messed up on my addition lol
haha around which part?
finding the antiderivative of u^(3/2)
okay I see..one dumb addition mistake lol
I dont think I can thank you enough tho, This most certainly helped me out tremendously, I def needed some help on this problem. Thank you again

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