anonymous
  • anonymous
integrate x(x-1)^(1/2) lower bound = 1 upper bound = 2 I know the answer but when I try and do it out I keep getting the wrong one, any help?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
myininaya
  • myininaya
\[\int\limits_{1}^{2}x \sqrt{x-1} dx\] Did you do u=x-1 => du =dx Also u=x-1 => u+1=x And if x=1 then u=1-1=0 And if x=2 then u=2-1=1 So we have \[\int\limits_{0}^{1} (u+1) \sqrt{u} du\]
myininaya
  • myininaya
Did you get this far?
myininaya
  • myininaya
oh and i like robots :)

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Shayaan_Mustafa
  • Shayaan_Mustafa
myininaya. i need your help. I messaged you. but you didn't reply me. :(
anonymous
  • anonymous
i cant see what you got because it says math processing answer
myininaya
  • myininaya
Try refreshing...
myininaya
  • myininaya
@imalittletoyrobotfromthefuture did that work?
anonymous
  • anonymous
yea it worked, i have that now. I didnt know what to do because I didnt recognize that x = u + 1
myininaya
  • myininaya
Well if u=x-1 then u+1=x (I just added 1 on both sides for u=x-1)
anonymous
  • anonymous
yes... haha i got that, what i had before though was: x\[\int\limits_{1}^{2}x \sqrt{u}du\]
myininaya
  • myininaya
But your integral needs to be in terms of one variable and when you make the switch from x to u your limits also need to switch from x to u
anonymous
  • anonymous
yes, thats why i was having trouble, i didnt recognize the fact that I could put x in terms of u, now that I have that though I'm unsure of what to do next
myininaya
  • myininaya
\[\int\limits\limits_{0}^{1} (u+1) \sqrt{u} du \] So this is what I wrote above
myininaya
  • myininaya
Do you understand exactly how I got it?
anonymous
  • anonymous
yes, what next though? this?: \[\int\limits_{0}^{1}((u^2/2)+u)((2u^3/2)/3)\]
myininaya
  • myininaya
?
anonymous
  • anonymous
isn't that the anti-derivitave of the equation \[\int\limits_{0}^{1}(u+1)\sqrt{u}du\]?
myininaya
  • myininaya
\[\int\limits_{0}^{1}(u+1) \sqrt{u} du=\int\limits_{0}^{1}(u+1)u^\frac{1}{2} du=\int\limits_{0}^{1}(u \cdot u^\frac{1}{2}+1 \cdot u^\frac{1}{2}) du\] \[=\int\limits_{0}^{1}(u^\frac{3}{2}+u^\frac{1}{2}) du\]
myininaya
  • myininaya
Now we integrate...
myininaya
  • myininaya
\[[\frac{u^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}]_0^1\]
myininaya
  • myininaya
\[=[\frac{1^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\frac{1^{\frac{1}{2}+1}}{\frac{1}{2}+1}]-[\frac{0^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\frac{0^{\frac{1}{2}+1}}{\frac{1}{2}+1}]\]
myininaya
  • myininaya
robot you have questions?
anonymous
  • anonymous
16/15 :) thanks for the explanation, very helpful and yes, robots are sick. check this out http://www.cnn.com/2012/03/04/opinion/ted-kumar-flying-robots/index.html?iref=obnetwork
myininaya
  • myininaya
lol

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