Using Integrals to find velocity?

Question

A car drives down a road in such a way that its velocity ( in m/s) at time t (seconds) is

v(t)= 3t^(1/2)+1

Find the car's average velocity between t=1 and t-3

Answer 1

v(1) = 3 + 1 = 4
v(3) = 3√3 + 1
(two readings)

Average reading = (v3 - v1) / 2 = (3√3 - 3) / 2 m/s
Average reading = 1.098 m/s

Answer 2

The velocity is v(t)
v(t)= 3t^(1/2)+1

Position is the integral of velocity.
V(t) = 3/(3/2) t^(3/2) + t + k
= 2t^(3/2) + t + k
To find average velocity between t=1 and t=3, we need to find out how far the car went in those 2 seconds between 1 and 3.
This would be V(3) - V(1)
= (13.39 + k) - (3+k)
= 10.39
The car traveled 10.39 m in 2 seconds.
Its average velocity during those two seconds was
10.39 / 2 = 5.2 m/s

Answer 3

I think the answer is by integrating the v(t) from 1-->3 then divide by (3-1)
(∫v(t) .dt)/(3-1) =(∫3t^0.5 +1)/2 = (2 t^(3/2) +t)/2 =t^3/2 +t/2
=3^3/2 +3/2 -1 -1/2 = 3^3/2 = 5.196152 m/s