## Calculus 8th Edition

$r(t)=(t^2+3t)i+(1-t) j+(\dfrac{t^3}{3}+1)k$
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. $r(t)=(t^2+3t)i+(1-t) j+(\dfrac{t^3}{3}+1)k$