anonymous
  • anonymous
ok, so now how would you integrate (5/2) i t ^ (3/2)? Same way? What happens to the i?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
The i stays put (it's a constant), assuming i is the imaginary number and you're integrating with respect to t. So you integrate normally here.
anonymous
  • anonymous
and 5/2 is a constant too :)
anonymous
  • anonymous
that's what I thought....okkkkk, so can you check my work out on this one? I had a really long integral problem and think I did all the simplifying corectly but I end up with (1/3) t^3 + (2/3) i t^(5/2) + i t^(3/2) between 0 and 1...do i just plug in the 1 and then the 0 for t and the i remains in the answer?

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anonymous
  • anonymous
Can you write out the question using the equation editor?
anonymous
  • anonymous
oh, how do I pull that up?
anonymous
  • anonymous
this is a new laptop and not sure what is on it exactly!!
anonymous
  • anonymous
(1/3) t^3 + (2/3) i t^(5/2) + i t^(3/2)
anonymous
  • anonymous
The equation editor is on the bottom of the message box. It says, "Equation"
anonymous
  • anonymous
\[\frac{1}{3}t^3+\frac{2}{3} i t^{5/2}+i t ^{3/2}\]...is that the thing you have to integrate or is that an answer to something?
anonymous
  • anonymous
actually, I have to intergrate the curve from (0,0) to (1,1) where y = sq rt of x using m(z) dz where m(z) is x^2 +y^2 + 2ixy
anonymous
  • anonymous
ok, I hit the eqn button but my tyoibng didnt come out like yours!
anonymous
  • anonymous
actually, the "contour" ...this is in Complex Analysis.
anonymous
  • anonymous
Yeah, I guessed. The path you have to integrate over is \[y=\sqrt{x}\]right?
anonymous
  • anonymous
and x = t
anonymous
  • anonymous
Yes, that's what I was going to suggest
anonymous
  • anonymous
and I THINK I did the distributive correctly...the dz/dt at the end of the original integral changes to (dx/dt) + i (dy/dt) and and i came up with integral of [t^2 + t +2i t^(3/2)] [ 1 + i (1/(2sqrt t)] dt
anonymous
  • anonymous
I double distributed and before simplifyling, came up with integral of t^2 + t + 2i t^(3/2) + i t^2 (1/(2sqrt t)) + i t (1/(2sqrt t) + 2i t^(3/2) (i) ((t^(1/2)/2)
anonymous
  • anonymous
could that be right for the distributive part??
anonymous
  • anonymous
just let me see
anonymous
  • anonymous
it's past my bedtime
anonymous
  • anonymous
oh, sorry...it's only past 10am here (east coast)
anonymous
  • anonymous
all of these fractional exponents and fractional constants are givng me a headache and making my eyes cross!
anonymous
  • anonymous
Were you actually given the function to integrate as\[\mu (z)=x^2+y^2+2i x y \]?
anonymous
  • anonymous
yes! :-)
anonymous
  • anonymous
Okay, you have \[z=x+iy\]as your variable and over this path, x and y will take values,\[x=t, y=t^{1/2}\]so\[z=z(t)=t+i t^{1/2}\]Your integral is\[\int\limits_{\gamma}^{}\mu (z) dz=\int\limits_{0}^{1}\mu(z(t))z'(t)dt\]where gamma is our defined path. Agree so far?
anonymous
  • anonymous
yep! that's sort of what the prof wrote...
anonymous
  • anonymous
Now, since mu(z) is as you defined it above in terms of x and y, and now that we have those x and y in terms of t along this contour, mu as a function of t is,\[\mu (t)=t^2+t+2i t^{3/2} \rightarrow \mu'(t)=2t+1+3 i t^{1/2} \]
anonymous
  • anonymous
oh forget mu'(t)...
anonymous
  • anonymous
late
anonymous
  • anonymous
\[z'(t)=1+i \frac{1}{2}t\]
anonymous
  • anonymous
woah....where did the 2 i t^ (3/2) come from?
anonymous
  • anonymous
So your integral becomes,\[\int\limits_{0}^{1}\left( t^2+t+2\it^{3/2} \right)\left( 1+i \frac{1}{2}t^{-1/2} \right) dt\]
anonymous
  • anonymous
yes, have that...
anonymous
  • anonymous
2it^(3/2) from subbing in x=t, y=t^(1/2) into 2ixy
anonymous
  • anonymous
Agreed?
anonymous
  • anonymous
got you, you were just wrtiting the original problem for me...thought htat had come from all the distributing...
anonymous
  • anonymous
The "i t" didn't show up for some reason in the integral above in the first bracket.
anonymous
  • anonymous
\[\int\limits_{0}^{1}\left( t^2+t+2 i t^{2/3} \right)\left( 1+i \frac{1}{2}t^{-1/2} \right)dt\]
anonymous
  • anonymous
That should be mistake-free
anonymous
  • anonymous
You have to expand that out and group the terms into an integral for the real and imaginary components, and then do Riemann integration ('normal' integration) on each interval.
anonymous
  • anonymous
\[\int\limits_{0}^{1}u(t)dt+i \int\limits_{0}^{1}v(t)dt\]
anonymous
  • anonymous
Happy so far?
anonymous
  • anonymous
yeah, I can finally type again...froze up or something
anonymous
  • anonymous
ok
anonymous
  • anonymous
ok, my question is, did I do the distributive math correctly?
anonymous
  • anonymous
You can check that with wolframalpha - have you heard of that?
anonymous
  • anonymous
way back up at 43 minutes ago... no, but I'll check it out...
anonymous
  • anonymous
Don't be surprised/put off if you distribute and get something rotten.
anonymous
  • anonymous
It happens all the time in complex analysis. It's only ever nice when the give you "Integrate z over the unit circle" or some pos.
anonymous
  • anonymous
actually, not bad... now can't find it, just had it
anonymous
  • anonymous
could I actually come up with something like t^2 + (3/2) i t ^(3/2) + (1/2 i t^ (1/2) before I integrate?
anonymous
  • anonymous
I got some like terms when I distributed so combined them
anonymous
  • anonymous
Before I answer that, again, there's a mistake in the last integral where I declared, "No mistakes"...it's late and this site is awkward. The power on the 2it in the first bracket should be 3/2, not 2/3.
anonymous
  • anonymous
You could come up with something like that before you integrate.
anonymous
  • anonymous
I'm double-checking my expansion.
anonymous
  • anonymous
sorry, didn't catch that...I hacethe exponent as 3/2
anonymous
  • anonymous
\[t^2+i \left( \frac{5}{2}t^{3/2}+\frac{1}{2}t^{1/2} \right)\]
anonymous
  • anonymous
You get 3/2, where I get 5/2
anonymous
  • anonymous
yeah! that's what I have !!!!!!! (just saw my mistake, yes the 3/2 should be a 5/2
anonymous
  • anonymous
So it's all good to go...
anonymous
  • anonymous
yes, so when integrating , I would get (1/3) t^3 for the real part and i [(t^5/2) + (1/3) t^(3/2) for the imaginary part?
anonymous
  • anonymous
That looks right.
anonymous
  • anonymous
Now sub your limits.
anonymous
  • anonymous
wahoo!@! so just plug in the 1 and 0 and evaluate as usual?
anonymous
  • anonymous
\[\frac{1}{3}+i \frac{4}{3}\]
anonymous
  • anonymous
so (1/3) + 2i???
anonymous
  • anonymous
Yes, you can evaluate as usual since you've converted a contour integral to one over an interval...so it's 'normal', for lack of a better term.
anonymous
  • anonymous
oos! wherer did your i (4/3) come from
anonymous
  • anonymous
Look at the imaginary part of your result. It's \[t^{5/2}+\frac{1}{2}t^{3/2}\]
anonymous
  • anonymous
sorry ^ 1/3
anonymous
  • anonymous
(1)^{5/2}+(1/3)*(1)^{3/2} = 3/3 + 1/3 = 4/3
anonymous
  • anonymous
fool thing froze again!! I saw my mistake while you were typing! duh, forgot to multiply in the 1/3
anonymous
  • anonymous
So you're fine now?
anonymous
  • anonymous
so I had (1/3)+ i + (1/3)i to get the two terms (1/3) + (4/3) i
anonymous
  • anonymous
yes
anonymous
  • anonymous
and when t=0, the result for the other limit is 0.
anonymous
  • anonymous
Sure am!! Now just have to do the same thing for contours y=x and y=x^2 and compare all 3 answers.
anonymous
  • anonymous
Well, good luck. It's the same deal :P
anonymous
  • anonymous
sorry, computer delayed again! THANK YOU soooooo MUCH FOR HELPING ME!! ;- )
anonymous
  • anonymous
No probs. I'm hoping you fanned me ;)
anonymous
  • anonymous
sure did!!!!! :-) and this site is free????
anonymous
  • anonymous
Yes
anonymous
  • anonymous
Okay, I'm off for sleep...have fun with complex analysis ;)
anonymous
  • anonymous
thanks, I will! :-)

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