Answer
$t^2i+t^3j+(\dfrac{2}{3}t^{3/2}-\dfrac{2}{3})k$
Work Step by Step
Given: $r(t)=\int (2ti+3t^2j+\sqrt t k)dt$
In order to evaluate the integral we will have to integrate each component of the function individually.
Let $I=\int (2ti+3t^2j+\sqrt t k)dt$
Thus,
$I=Ai+ Bj+Ck$ ... (1)
Here, $A=2t$ and $B=3t^2, C=\sqrt t $
Consider $A=2t$
Thus, $A=2t=2(\frac{1}{2})t^2+C$
Now, consider $B=3t^2=3(\frac{1}{3})t^3+C'$
and $C=\sqrt t =(\frac{1}{3/2})t^{3/2}+C''$
Plug in the values of A,B and C in equation (1) and we get
$I=(2(\frac{1}{2})t^2+C)i+(3(\frac{1}{3})t^3+C')j+((\frac{1}{3/2})t^{3/2}+C'')k$
Here, C,C',C'' are the constants of integration.
As we are given $r(1)=i+j$ which can be written as $\lt 1,1,0 \gt$.
Therefore, $C=0,C'=0,C''=-\dfrac{2}{3}$
Hence,
$r(t)=(2(\frac{1}{2})t^2+0)i+(3(\frac{1}{3})t^3+0)j+((\frac{1}{3/2})t^{3/2}+-\dfrac{2}{3})k$
or,
$r(t)=t^2i+t^3j+(\dfrac{2}{3}t^{3/2}-\dfrac{2}{3})k$
