## Thomas' Calculus 13th Edition

$\dfrac{1}{2}$
The work done can be found as: $W=\int_a^b F(r(t)) \dfrac{dr}{dt}(dt) ...(1)$ Here, $\dfrac{dr}{dt}=i+2t j+k$ Equation (1) becomes: $W=\int_0^{1} (t^3 i+t^2 t j-t^3 k)\cdot (i+2t j+k) dt \\=\int_0^1 t^3+2t^3 -t^3 \ dt \\ =\int_0^1 2t^3 \ dt \\= \dfrac{1}{2} \times [t^4]_0^1 \\=\dfrac{1}{2}$