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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
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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In 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 \(dx=\dfrac{\sec^2t}{\tan t}\,dt=(\cot t+\tan t)\,dt\). \[\begin{align*}L&=\int_{\pi/4}^{\arctan e^2}\sqrt{1+\tan^2t}\,(\cot t+\tan t)\,dt\\\\ &=\int_{\pi/4}^{\arctan e^2}\sec t(\cot t+\tan t)\,dt\\\\ &=\int_{\pi/4}^{\arctan e^2} \left(\csc t+\sec t\tan t\right)\,dt \end{align*}\]
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