Here's the question you clicked on:
rosho
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
do you want to me to draw the equation?
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