CalcII how to solve the arc length: a) int{a,b}sqrt(1+16x^4) b) int{a,b}sqrt(1+36cos^2(2x)) ans a)f(x)=+/-4x^3/3+C b)+/-3sin2x+C

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CalcII how to solve the arc length: a) int{a,b}sqrt(1+16x^4) b) int{a,b}sqrt(1+36cos^2(2x)) ans a)f(x)=+/-4x^3/3+C b)+/-3sin2x+C

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do you want to me to draw the equation?
|dw:1363864864886:dw|

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start with that problem, i'll draw the second when someone has explained the answer.
Are you sure you entered that correctly..? This is the solution Wolfram is giving, http://www.wolframalpha.com/input/?i=integral+sqrt%281%2B16x%5E4%29dx There must be a mistake somewhere in there.
we are trying to find the arc length. i think.
So find the arclength of this function, \(\large f(x)=\sqrt{1+16x^4}\) From \(\large a\) to \(\large b\) ?
i'll type the text verbatim from the book. what differentiable function have an arc length on the interval [a,b] given by the following integrals?
Oh oh oh ok I see what's going on.
okay, that's good because i have no idea.
The formula for arc length is,\[\large \int\limits ds \qquad = \qquad \int\limits \sqrt{1+\color{orangered}{\left(\frac{dy}{dx}\right)^2}}dx\] And we're given this,\[\large \int\limits \sqrt{1+\color{orangered}{16x^4}}dx\]
oh, i get it now... thanks a bunch

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