the answer is (c)
Since position vector r(t)=x(t)i+y(t)j+z(t)kr(t)=x(t) i+y(t)j+z(t)kr(t)=x(t)i+y(t)j+z(t)k
and from the hypothesis that r(t)=0=0i+0j+0kr(t)=0=0 i+0j+0kr(t)=0=0i+0j+0k
Therefore ∫r(t)dt=∫(0i+0j+0k)dt\int r(t)dt=\int( 0 i+0j+0k )dt∫r(t)dt=∫(0i+0j+0k)dt
and then
∫r(t)dt=c1i+c2j+c3k=c,\int r(t)dt=c_{1} i+c_{2}j+c_{3}k=c,∫r(t)dt=c1i+c2j+c3k=c,
where ccc is an arbitrary constant vector
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