compute thearc length of the curve f(x)= ln(1+x^2) -1/16 x^2 -1/8 ln(x) <a> ln 9 +3/16 +7/8 ln2 <b> ln 5 +3/16 -7/8 ln2 <c> ln 9 +3/16 <d> ln5- 3/16
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Originally Posted by harry compute thearc length of the curve f(x)= ln(1+x^2) -1/16 x^2 -1/8 ln(x) <a> ln 9 +3/16 +7/8 ln2 <b> ln 5 +3/16 -7/8 ln2 <c> ln 9 +3/16 <d> ln5- 3/16 Aren't there some limits involved here? RonL
compute thearc length of the curve f(x)= ln(1+x^2) -1/16 x^2 -1/8 ln(x) limits of x are [1,2] <a> ln 9 +3/16 +7/8 ln2 <b> ln 5 +3/16 -7/8 ln2 <c> ln 9 +3/16 <d> ln5- 3/16
Last edited by harry; March 18th 2007 at 12:55 PM.
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