parametize C of the line integral as
[tex]({e}^{t} ,t)[/tex]
from 0<=t<1 then
[tex]dx = {e}^{t} dt[/tex]
which means by substituting x with e^t and y with and dx with e^t dt we get
[tex]integral \: from \: 0 \: to \: 1( {e}^{t} \times {e}^{t} \times {e}^{t} \: dt )[/tex]
by adding exponents
[tex]integrl \: from \: 1 \: to \: 0 \: of \: ( {e}^{3t}dt )[/tex]
[tex] \frac{ {e}^{3} - 1}{3} - \frac{ {e}^{0} - 1 }{3} [/tex]
