Answer
$(\dfrac{1}{2}t^2+1) i+t^2j+tk$
Work Step by Step
As we are given that $a(t)=i+2j$
Since, $v(t)=\int a(t)$ and $r(t)=\int v(t)$
Thus, $v(t)=\int (i+2j) dt$
$\implies v(t)=ti+2tj+k$
Now, $r(t)=\int [ti+2tj+k] dt=\dfrac{1}{2}t^2i+t^2j+tk+ c$
Here, $c$ represents a constant of integration.
That is, $r(0)=i \implies c= i$
or, $r(t)=\dfrac{1}{2}t^2i+t^2j+tk+i$
Hence, $r(t)=(\dfrac{1}{2}t^2+1) i+t^2j+tk$
