Asked by Heather
Find the arclength of y=x^3/3+1/(4x) from 1 to 2. I know that arclength is the integral from 1 to 2 of the sqrt of (1 + (y')^2)dx, but have not been able to figure out how to evaluate the integral. Your help is much appreciated!
Answers
Answered by
Steve
1+y'^2 = 1+(x^2-1/(4x^2))^2
= (x^2 + 1/(4x^2))^2
so,
sqrt(1+y'^2) = x^2 + 1/(4x^2)
that should be easy to handle
= (x^2 + 1/(4x^2))^2
so,
sqrt(1+y'^2) = x^2 + 1/(4x^2)
that should be easy to handle
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