Calc II, finding the velocity and distanced travel

A particle is moving along a straight line and the acceleration function (in m/s^2) is a(t)=2t+3.

a)Find the inital velocity at time t, if the inital velocity was v(0)= -10m/s.
so i got v(t) = t^2 + 3t -10

b) find the total distance traveled during the time interval [0,3]

(a.) a(t) = 2t + 3

v(t) = t^2 + 3t + C

-10 = C

v(t) = t^2 + 3t - 10

(b.) Total distance = -∫(t² + 3t - 10) dt from 0 to 1 + ∫(t² + 3t - 10) dt from 1 to 3

= -t^3/3 - (3t^2)/2 + 10t eval. from 0 to 1 + (t^3/3 + 3t^2/2 - 10t) eval. from 1 to 3

= -1/3 - 3/2 + 10 + 9 + 27/2 - 30 - 1/3 - 3/2 + 10 = 8.83 m

part (a) is correct.

part (b) : by the Fundamental Theorem of Calculus (call this theorem FTC) , F(b) - F(a) = integral f(x)*dx [a,b] , if F ' = f

Call s(t) = displacement or distance function... so that s ' (t) = v (t)... rate of change of distance over time is velocity.

So, by the FTC above, s (3) - s (0) = integral of ( t^2 + 3t -10 ) over 0 < = t < = 3. This is the total distance traveled during this time interval.

The right side is t^3 / 3 + 3t^2 / 2 - 10t (sub t = 3 and sub t=0 , and subtract) = -7.5 m

answer: = -7.5 m