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Calculus1
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arc length of y=(1/4)x^2-(1/2)ln(x) from x=1 to 2
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these problems are always cooked up to be doable did you find \(y'\)?
i think it is \[y'=\frac{1}{2}x-\frac{1}{2x}\]
then you have to square it
something like \[\frac{(x^2-1)^2}{4 x^2}\]
then you have to add one to it
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and that is where the miracle occurs you get \[\frac{(x^2+1)^2}{4 x^2}\] which is a perfect square
good thing since now you have to take the square root of it that gives you \[\int_1^2\frac{x^2+1}{2x}dx\]
imagine how much work went in to making this problem work out so nicely i tried it once took me over half an hour to make one
@satellite73 i did the same way as u said to calculate Arithmetic mean(before asking)
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