Answer
$\dfrac{6}{5}-\cos 1 -\sin 1$
Work Step by Step
Here, $dr=(3t^2i-2tj+k) dt$ and $F(r(t)=\sin t^3i +\cos t^2 j+t^4 k$
$\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}=\int_0^{1} (\sin t^3i +\cos t^2 j+t^4 k) \cdot (3t^2 i -2tj+k) d t= \int_0^{1} 3t^2 \sin t^3-2t \cos t^2+t^4 dt$
or, $=[-\cos t^3-\sin t^2+\dfrac{t^5}{5}]_0^1$
or, $\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}=\dfrac{6}{5}-\cos 1 -\sin 1$
