Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 17 - Line and Surface Integrals - 17.2 Line Integrals - Exercises - Page 935: 73

Answer

The average value of $f$ along the curve is $Av\left( f \right) \simeq 0.574$.

Work Step by Step

The curve $C$: $y = {x^2}$ for $0 \le x \le 1$, can be parametrized by ${\bf{r}}\left( t \right) = \left( {t,{t^2}} \right)$, ${\ \ \ }$ for $0 \le t \le 1$ So, $||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {1,2t} \right)\cdot\left( {1,2t} \right)} = \sqrt {1 + 4{t^2}} $ The length of $C$ is $L = \mathop \smallint \limits_C^{} {\rm{d}}s = \mathop \smallint \limits_0^1 ||{\bf{r}}'\left( t \right)||{\rm{d}}t$ $L = \mathop \smallint \limits_0^1 \sqrt {1 + 4{t^2}} {\rm{d}}t$ From the Table of Integrals at the end of the book we know that $\smallint \sqrt {{a^2} + {u^2}} {\rm{d}}u = \frac{u}{2}\sqrt {{a^2} + {u^2}} + \frac{{{a^2}}}{2}\ln \left( {u + \sqrt {{a^2} + {u^2}} } \right) + C$ So, $L = \mathop \smallint \limits_0^1 \sqrt {1 + 4{t^2}} {\rm{d}}t = \frac{1}{2}\mathop \smallint \limits_0^1 \sqrt {1 + {{\left( {2t} \right)}^2}} {\rm{d}}\left( {2t} \right)$ $L = \frac{1}{2}\left[ {\frac{{2t}}{2}\sqrt {1 + 4{t^2}} |_0^1 + \frac{1}{2}\ln \left( {2t + \sqrt {1 + 4{t^2}} } \right)|_0^1} \right]$ $L = \frac{1}{2}\left[ {\sqrt 5 + \frac{1}{2}\ln \left( {2 + \sqrt 5 } \right)} \right]$ Using Eq. (11) we calculate the average value of $f\left( {x,y} \right) = x$ along the segment $C$ as $Av\left( f \right) = \frac{1}{L}\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s$ By Eq. (4), the integral becomes $Av\left( f \right) = \frac{1}{L}\mathop \smallint \limits_0^1 f\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$ $Av\left( f \right) = \frac{1}{L}\mathop \smallint \limits_0^1 t\sqrt {1 + 4{t^2}} {\rm{d}}t$ Write $u = 1 + 4{t^2}$, $du = 8tdt$. So, $Av\left( f \right) = \frac{1}{{8L}}\mathop \smallint \limits_0^1 {u^{1/2}}{\rm{d}}u$ $Av\left( f \right) = \frac{1}{{12L}}{u^{3/2}}|_1^5 = \frac{1}{{12L}}\left( {5\sqrt 5 - 1} \right)$ Substituting $L = \frac{1}{2}\left[ {\sqrt 5 + \frac{1}{2}\ln \left( {2 + \sqrt 5 } \right)} \right]$ in $Av\left( f \right)$, gives $Av\left( f \right) = \frac{1}{{6\sqrt 5 + 3\ln \left( {2 + \sqrt 5 } \right)}}\left( {5\sqrt 5 - 1} \right) \simeq 0.574$ So, the average value of $f$ along the curve is $Av\left( f \right) \simeq 0.574$.
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