Answer
$r(t)=(2(\frac{1}{2})t^2+0)i+(3(\frac{1}{3})t^3+0)j+((\frac{1}{3/2})t^{3/2}-\dfrac{2}{3})k$
or,
$r(t)=t^2i+t^3j+(\dfrac{2}{3}t^{3/2}-\dfrac{2}{3})k$
Work Step by Step
Suppose $I=\int (2ti+3t^2j+\sqrt t k)dt$
Rewrite this equation as: $I=Ai+ Bj+Ck$ ... (a)
Here, $P=2t$ and $Q=3t^2, R=\sqrt t $
$P=2t=2(\frac{1}{2})t^2+C$
and $Q=3t^2=3(\frac{1}{3})t^3+C'$
and $R=\sqrt t =(\frac{1}{3/2})t^{3/2}+C''$
From equation (a):
$I=(2(\frac{1}{2})t^2+C)i+(3(\frac{1}{3})t^3+C')j+((\frac{1}{3/2})t^{3/2}+C'')k$;(C,C',C'' are the constant of integration.)
From the question, $r(1)=i+j$ gives $\lt 1,1,0 \gt$.
After simplifications, we get: $C=0,C'=0,C''=-\dfrac{2}{3}$
The desired result is:
$r(t)=(2(\frac{1}{2})t^2+0)i+(3(\frac{1}{3})t^3+0)j+((\frac{1}{3/2})t^{3/2}-\dfrac{2}{3})k$
or,
$r(t)=t^2i+t^3j+(\dfrac{2}{3}t^{3/2}-\dfrac{2}{3})k$
