## anonymous one year ago Find the arc length by integration for y=e^x , y=e^-x x from 0 to 2 please help here

1. anonymous

So you're looking for the arc length of two different curves over the same interval, or are you looking at a sort of closed loop?|dw:1433945291686:dw|

2. anonymous

In either case, you can use the arc length integral formula. For a curve $$C$$ defined by a differentiable function $$y=f(x)$$ defined over an interval $$[a,b]$$, the arc length is given by $\int_CdS=\int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$ Consider the first curve: $$y=e^x$$, so $$\dfrac{dy}{dx}=e^x$$. The arc length $$L$$ is then $L=\int_0^2\sqrt{1+e^{2x}}\,dx$ You can compute the integral using a trigonometric substitution, $$e^x=\tan t$$ such that $$e^x\,dx=\sec^2t\,dt$$, or $$dx=\dfrac{\sec^2t}{\tan t}\,dt=(\cot t+\tan t)\,dt$$. \begin{align*}L&=\int_{\pi/4}^{\arctan e^2}\sqrt{1+\tan^2t}\,(\cot t+\tan t)\,dt\\\\ &=\int_{\pi/4}^{\arctan e^2}\sec t(\cot t+\tan t)\,dt\\\\ &=\int_{\pi/4}^{\arctan e^2} \left(\csc t+\sec t\tan t\right)\,dt \end{align*}