Answer
$r(t)=(t^2+3t)i+(1-t) j+(\dfrac{t^3}{3}+1)k$
Work Step by Step
As we know $v(t)=\int a(t)$ and $r(t)=\int v(t)$
This yields, $v(t)=\int (2i+2tk) dt$
or, $ v(t)=(2t+3)i-j+t^2k$
and $r(t)=\int [(2t+3)i-j+t^2k]dt=(t^2+3t)i-t j+\frac{t^3}{3}k+c$
Also, $r(0)=j+k$ and $ c= j+k$
Thus, $r(t)=\int [(2t+3)i-j+t^2k]=(t^2+3t)i-t j+\frac{t^3}{3}k+j+k$
Hence, the result.
$r(t)=(t^2+3t)i+(1-t) j+(\dfrac{t^3}{3}+1)k$
