Question

Value of the Line integral of xe^y dx over arc of x=e^y from pt (1,0) to (e,1). Please don't bother if you don't know how to do it. Thanks

Answer 1

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