How does one find the length of y = x^3/3 + 1/(4x) from x = 1 to 2?
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Originally Posted by Unenlightened How does one find the length of y = x^3/3 + 1/(4x) from x = 1 to 2? The formula for arc length is $\displaystyle s= \int_{a}^{b}\sqrt{a+f'[x]^2}dx$, where the arc runs from a to b. Look familiar?
Originally Posted by Unenlightened How does one find the length of y = x^3/3 + 1/(4x) from x = 1 to 2? arclenght is given by $\displaystyle \int_a^{b}\sqrt{1+f'(x)^2}dx$ so in this case it would be $\displaystyle \int_1^{2}\sqrt{1+\bigg(x^2+\frac{-1}{(4x)^2}\bigg)^2}dx$
Cheers - it's not actually for me. A friend was asking me and I couldn't remember how to do it. We don't use numbers anymore in my course
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