Answer
The work done by the field along the given path is $2{\rm{e}} - \frac{{39}}{{20}}$.
Work Step by Step
We obtain
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {{{\rm{e}}^{{t^2}}},{{\rm{e}}^t},\frac{{{t^4}}}{2}} \right)$
${\bf{r}}'\left( t \right) = \left( {2t,1,\frac{1}{2}} \right)$
The work done by the field:
$W = \mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^1 {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$W = \mathop \smallint \limits_0^1 \left( {{{\rm{e}}^{{t^2}}},{{\rm{e}}^t},\frac{{{t^4}}}{2}} \right)\cdot\left( {2t,1,\frac{1}{2}} \right){\rm{d}}t$
$ = \mathop \smallint \limits_0^1 \left( {2t{{\rm{e}}^{{t^2}}} + {{\rm{e}}^t} + \frac{{{t^4}}}{4}} \right){\rm{d}}t$
$ = \mathop \smallint \limits_0^1 {{\rm{e}}^{{t^2}}}{\rm{d}}\left( {{t^2}} \right) + \mathop \smallint \limits_0^1 {{\rm{e}}^t}{\rm{d}}t + \frac{1}{4}\mathop \smallint \limits_0^1 {t^4}{\rm{d}}t$
$ = \left( {{{\rm{e}}^{{t^2}}}|_0^1 + {{\rm{e}}^t}|_0^1 + \frac{1}{{20}}{t^5}|_0^1} \right)$
$ = {\rm{e}} - 1 + {\rm{e}} - 1 + \frac{1}{{20}} = 2{\rm{e}} - \frac{{39}}{{20}}$
So, the work done by the field along the given path is $2{\rm{e}} - \frac{{39}}{{20}}$.
