anonymous
  • anonymous
I cannot find where I messed up in this problem. Find the exact arc length of y=sqrt(x-x^2)+sin^-1(sqrt(x))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
from 0 to 1 i take it
anonymous
  • anonymous
you need \[\int_0^1\sqrt{1+f'(x)^2}dx\] if i recall correctly, so first you need \[f'(x)=\frac{\sqrt{x-x^2}}{x}\]
anonymous
  • anonymous
and so \[f'(x)^2=\frac{x-x^2}{x^2}=\frac{1}{x}-x\]

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More answers

anonymous
  • anonymous
add 1 and get \[\frac{-x^2+x+1}{x}\]
anonymous
  • anonymous
damn that was wrong \[f'(x)^2=\frac{1}{x}-1\] so \[1+f'(x)^2=\frac{1}{x}\]
anonymous
  • anonymous
now you have \[\int_0^1\sqrt{\frac{1}{x}}dx\] which should be ok right?
anonymous
  • anonymous
no, there were not any intergration limits
anonymous
  • anonymous
It wanted the exact value
anonymous
  • anonymous
yea I got that
anonymous
  • anonymous
but when I put in 2sqrt(x) as my answer, it said it was wrong
anonymous
  • anonymous
well the funciton is only defined on the interval [0,1] so i assumed those were the limits of integration
anonymous
  • anonymous
oh okay, I just assumed it meant no limits
anonymous
  • anonymous
integral is 2
anonymous
  • anonymous
i guess when it said "exact arc length" it meant the total arc length
anonymous
  • anonymous
Haha, yay I got it. Thank for your help!!! Yea, I did not exactly understand that part. But it is all good now.
anonymous
  • anonymous
great! yw

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