Answer
work done $\int_C \textbf F \cdot \textbf{dr} = 66$ units
Work Step by Step
The path is given by
$C: x=t, y = t^3$ from (0,0) to (2,8)
This path can be written as
$ \textbf r = x\textbf i + y\textbf j \\
= t \textbf i + t^3 \textbf j, \text{for } 0\le t \le 2\\
\therefore \textbf r' = \textbf i + 3 t^2 \textbf j$
Given force field $\textbf F = x \textbf i +2y \textbf j\\
=t \textbf i +2 t^3 \textbf j$
work done = $\int_C \textbf F \cdot d\textbf r = \int _{t=0}^{2 }\textbf F \cdot \textbf r' dt\\
= \int _{0}^{2 }(t \textbf i +2 t^3 \textbf j)\cdot (\textbf i + 3 t^2 \textbf j) dt\\
= \int _{0}^{2 } [t+6t^5]dt\\
=[\frac{t^2}{2}+t^6]_0^{2}\\
= 66$
workdone = $66$ units
