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 $Av\left( f \right) \simeq 0.574$.