Answer
$0.5424$
Work Step by Step
Here, we have $F[r(t)]=e^{t-t^2} i +\sin (e^{-t^2}) j$
and $dr=(e^ti-2te^{-t^2} j) dt$
$\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}=\int_1^{2} (e^{t-t^2} i +\sin (e^{-t^2}) j) \cdot (e^ti-2te^{-t^2} j) dt$
This implies that
$\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}= \int_1^{2} (e^{2t-t^2}-2t (e^{-t^2}) \sin (e^{-t^2}) dt$
Need to use calculator.
$\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}=\int_1^{2} (e^{2t-t^2}-2t (e^{-t^2}) \sin (e^{-t^2}) dt=0.5424$
