anonymous
  • anonymous
How do you take the antiderivative of (2x-1)^-1/2???
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Is that (2x-1)^(-1/2) or (2x-1)^-1 which is divided by 2?
anonymous
  • anonymous
(2x-1)^(-1/2)
anonymous
  • anonymous
O.k. Then you would let u = 2x-1 Then du = 2 dx So your integral:\[\int\limits \frac{1}{\sqrt{2x+1}}dx\] becomes

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anonymous
  • anonymous
1/2 times \[\int\limits \frac{2dx}{\sqrt{2x+1}}\]
anonymous
  • anonymous
\[\int\limits \frac{du}{\sqrt{u}}=\]
anonymous
  • anonymous
\[\int\limits u ^{-1/2}du=\]
anonymous
  • anonymous
\[\frac{u ^{-1/2 + 1}}{-1/2+1}+C\]
anonymous
  • anonymous
\[\frac{(2x+1) ^{1/2}}{1/2}+C\]
anonymous
  • anonymous
and dividing by 1/2 is the same as multiplying by the reciprocal 2/1
anonymous
  • anonymous
What about x/(2x-1)^(1/2) with an upper of 5 and a lower of 1. My calculator is giving me a correcct answer of 5.333333. But I can work the problem out to be the same
anonymous
  • anonymous
sorry, I "can't" work the problem out the same
anonymous
  • anonymous
o.k. well, you would want to plug 5 into \[2*\sqrt{2x+1}\] and then subtract what you get when you plug 1 into it.
anonymous
  • anonymous
that would be 2 times the result of \[\sqrt{2(5)+1} - \sqrt{2(1)+1} = \sqrt{11}-\sqrt{3}\]
anonymous
  • anonymous
which turns out to be about 3.17
anonymous
  • anonymous
Oh! I didn't notice you had an "x" in the numerator.
anonymous
  • anonymous
You would still use the same substitution
anonymous
  • anonymous
but the x in the numerator can be rewritten in terms of "u"
anonymous
  • anonymous
right. x= (1+u)/2
anonymous
  • anonymous
since u = 2x+1 u-1 = 2x and so you get that x = (.5)*(u - 1)
anonymous
  • anonymous
so dividing that by \[\sqrt{u} = u^{1/2}\] you can use exponent rules to get
anonymous
  • anonymous
\[\frac{.5*(u-1)}{u^{1/2}} \]
anonymous
  • anonymous
which boils down to \[.5* (u^{1-1/2}-u^{-1/2})\]
anonymous
  • anonymous
which is\[.5* (u^{1/2}-u^{-1/2})\]
anonymous
  • anonymous
and you can use the power rule for integrals to antidifferentiate
anonymous
  • anonymous
\[.5* (u^{3/2}-u^{1/2})\]
anonymous
  • anonymous
or \[.5* ((2x+1)^{3/2}-(2x+1)^{1/2})\]
anonymous
  • anonymous
does that make sense?
anonymous
  • anonymous
Yes, but when I plug in the upper and lower I'm still not getting the correct answer.......
anonymous
  • anonymous
let me check
anonymous
  • anonymous
The (2x+1) is suppose to be (2x-1)
anonymous
  • anonymous
right...it's 14.8511...
anonymous
  • anonymous
Oh! :) Well, I can guarantee you if you use 2x-1 instead it should work but I'll check
anonymous
  • anonymous
My graphing calculator is giving me 5.333333
anonymous
  • anonymous
\[9^{3/2}-9^{1/2}-(1^{3/2}-1^{1/2})\]
anonymous
  • anonymous
is \[3\sqrt{3}-3\]
anonymous
  • anonymous
Do you have a graphing cal on you?
anonymous
  • anonymous
Actually, you have to go back and change the substitution also so it is more work than this
anonymous
  • anonymous
if u = 2x-1 then x=.5*(u+1)
anonymous
  • anonymous
Thanks for your help. I'm in FLorida and it's about 1am. I need to hit the hay. Thanks again......
anonymous
  • anonymous
Yeah, I did it by hand and it works out. Just use that substitution
anonymous
  • anonymous
and also multiply by the reciprocal of the new exponent you get when you integrate. ;)
anonymous
  • anonymous
Will do. Thanks!
anonymous
  • anonymous
Your welcome...sorry I'm new to typing this out. It's a bit distracting ;) but I'll get used to it. Night!
anonymous
  • anonymous
Yeah, it's a pain in the retrice Laters

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