Answer
$$\int_CF.dr=\frac{11e-48}{12e}$$
Work Step by Step
$\int_CF.dr=\int_0^{1}2te^{-t}dt+\int_0^{1}-3t^5-t^2-t^3dt$
$=[-2te^{-t}-2e^{-t}]_0^{1}+[-\frac{t^6}{2}-\frac{t^3}{3}-\frac{t^4}{4}]_0^{1}$
Let $A=[-2te^{-t}-2e^{-t}]_0^{1}=2-(\dfrac{4}{e})$
Let $B=[-\frac{t^6}{2}-\frac{t^3}{3}-\frac{t^4}{4}]_0^{1}=-\dfrac{13}{12}$
$=2-(\dfrac{4}{e})-\dfrac{13}{12}$
$A+B=\dfrac{24e-48-13e}{12e}$
$$\int_CF.dr=\frac{11e-48}{12e}=\frac{11}{12}-\frac{4}{e}$$
