Answer
$0.5424$
Work Step by Step
Here, we have $dr=(e^ti-2te^{-t^2} j) dt$ and $F(r(t))=e^{t-t^2} i +\sin (e^{-t^2}) j$
The line integral is given by: $\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}=\int_1^{2} (e^{t-t^2} i +\sin (e^{-t^2}) j) \cdot (e^ti-2te^{-t^2} j) dt$
or, $\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}= \int_1^{2} e^{2t-t^2}-2t (e^{-t^2}) \sin (e^{-t^2}) dt$
Now using the calculator, we get
$\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}=0.5424$
