Evaluate the line integral along the path C given by

Evaluate the line integral along the path C given by where $0\le t\le 1\int cxydx+ydy$
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Momp1989
${\int }_{C}xydx+ydy$
Given that $x=2t$ and $y=4t$
Therefore, $dx=2dt$ and $dy=4dt$
Substitute the value to get
${\int }_{C}\left(2t×4t\right)\left(2dt\right)+\left(4t\right)\left(4dt\right)$
Given that $0\le t\le 1$, therefore we will integrate from 0 to 1
${\int }_{0}^{1}16{t}^{2}+16tdt={\left[\frac{16{t}^{3}}{3}+\frac{16{t}^{2}}{2}\right]}_{0}^{1}$
$={\left[\frac{16{t}^{}}{3}+8{t}^{2}\right]}_{0}^{1}$
$=\frac{16}{3}+8$
$=\frac{16+24}{3}$
$=\frac{40}{3}$