Answer
The distance the particle moved after time t is
$$x(t)=-e^{-t}(t^{2}+2t+2) meter$$
Work Step by Step
the distance that the particle moved is a function of time and it's the integral of the velocity function with respect to time
so the the distance moved is
$$x(t)=\int{v(t)} dt=\int{t^{2}e^{-t}}dt$$
which can be solved by integration by parts using the formula $A=uv-\int{vdu}$
$$u=t^{2}$$
and
$$dv=e^{-t}$$
so applying the formula the integral becomes
$$A=-t^{2}e^{-t} +\int{2te^{-t}}dt$$
the second term is also an integral that can be solved by parts
$$\int{te^{-t}}dt=- te^{-t}+\int{e^{-t}}dt= -te^{-t}- e^{-t}$$
so
$$A=-t^{2}e^{-t}-2te^{-t}-2 e^{-t}$$
the distance is
$$x(t)=-e^{-t}(t^{2}+2t+2) m$$
