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## Homework Statement

Find the arclength of the function [tex] y=x^2 [/tex] when x is between 0 and 10.

## Homework Equations

Arclength here is [tex] \int_{0}^{10} \sqrt{1+(2x)^2} dx [/tex]

(It's intended to be the integral from 0 to 10 of the quare root of 1+(2x)^2. My latex skills suck.)

## The Attempt at a Solution

Trig substitution. 2x= tan (theta)

Integrate secant theta

Which is ln(sec(theta)+tan(theta))

Substituting theta back in for arcsin(2x)

ln(sec(arctan(2x))+tan(arctan(2x))) evaluated from 10 to 0. Solving for arctan(2x) my final answer is

ln(sqrt(1+4x^2)+2x) evaluated from 0 to 10. I know this is wrong as it is much to short of an arc length but I don't know where I went wrong. Any clues?

## The Attempt at a Solution

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