Question

26 20: Find the cray length of the curvey from =0 tox= 1. Show all of the work in detail. You will be graded on setting up the correct integral and then showing the steps using our integration techniques to find an ili derivative of the in legrand which will allow you to evaluate the integral.

Answer 1

Given f(1) => 1 = 0 to 2 = 1

$Length \:of \:curve \:L \:= \int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}\:dx$

$L \:= \int_{0}^{1}\sqrt{1+(\frac{dy}{dx})^2}\:dx$

$\frac{dy}{dx}=\frac{\mathrm{d} (\frac{x^2}{2})}{\mathrm{d} x}=\frac{2x}{2}=x$

$L \:= \int_{0}^{1}\sqrt{1+(x)^2}\:dx$

Apply Trig Substitution : x = tan (u)

$\frac{dx}{du}=\sec ^2\left(u\right)$

$\Rightarrow \:dx=\sec ^2\left(u\right)du$

$=\int \sqrt{1+\tan ^2\left(u\right)}\sec ^2\left(u\right)du$

Simplify V1+tan? (u) sec (u)

$\rightarrow secu*sec^2u=sec^3u$

$Therefore \:integral \:converts \:to =\int \sec ^3\left(u\right)du$

Adjust integral boundaries :

$\:x=0\quad \Rightarrow \:u=0$

$\:x=1\quad \Rightarrow \:u=\frac{\pi }{4}$

$=\int _0^{\frac{\pi }{4}}\sec ^3\left(u\right)du$

$\mathrm{Apply\:Integral\:Reduction}:\quad \int \sec ^n\left(x\right)dx=\frac{\sec ^{n-1}\left(x\right)\sin \left(x\right)}{n-1}+\frac{n-2}{n-1}\int \sec ^{n-2}\left(x\right)dx$

$\int \sec ^3\left(u\right)du=\frac{\sec ^2\left(u\right)\sin \left(u\right)}{2}+\frac{1}{2}\int \sec \left(u\right)du$

$=\left[\frac{\sec ^2\left(u\right)\sin \left(u\right)}{2}\right]^{\frac{\pi }{4}}_0+\frac{1}{2}\cdot \int _0^{\frac{\pi }{4}}\sec \left(u\right)du --------- (1)$

$Now\:Evaluate \int _0^{\frac{\pi }{4}}\sec \left(u\right)du$

$\mathrm{Use\:the\:common\:integral}:\quad \int \sec \left(u\right)du=\ln \left|\tan \left(u\right)+\sec \left(u\right)\right|$

$=\left[\ln \left|\tan \left(u\right)+\sec \left(u\right)\right|\right]^{\frac{\pi }{4}}_0$

$\left[\ln \left|\tan \left(u\right)+\sec \left(u\right)\right|\right]^{\frac{\pi }{4}}_0=\ln \left(1+\sqrt{2}\right)$

$=\left[\frac{\sec ^2\left(u\right)\sin \left(u\right)}{2}\right]^{\frac{\pi }{4}}_0+\frac{1}{2}\ln \left(1+\sqrt{2}\right)$

$we \:know\: \frac{\sec ^2\left(u\right)\sin \left(u\right)}{2}=\frac{\sec \left(u\right)\tan \left(u\right)}{2}$

$=\left[\frac{\sec \left(u\right)\tan \left(u\right)}{2}\right]^{\frac{\pi }{4}}_0+\frac{1}{2}\ln \left(1+\sqrt{2}\right)$

$=\frac{1}{\sqrt{2}}+\frac{1}{2}\ln \left(1+\sqrt{2}\right)$

$=\frac{\sqrt{2}+\ln \left(1+\sqrt{2}\right)}{2}$

$Therefore \:Length \:of \:curve \:=\frac{\sqrt{2}+\ln \left(1+\sqrt{2}\right)}{2}$