Determining the displacement of a particle from velocity function by integration

A particle moves along a straight line from a fixed point $\displaystyle A$.Its velocity,

$\displaystyle v m s^-1,t s$ after leaving point $\displaystyle A$ is given by $\displaystyle v=3 + 2t-t^2$

a)Find the total distance travelled by the particle between $\displaystyle t=2$ and $\displaystyle t=4$

i get,

total distance travelled by the particle between $\displaystyle t=2$and $\displaystyle t= 3$

$\displaystyle =\int_{2}^{3} v dt$

$\displaystyle =\int_{2}^{3} (3 + 2t - t^2) dt$

$\displaystyle =\left[3t+t^2-\frac{t^3}{3} \right]_{2}^{3}$

$\displaystyle =\left[3(3) + 3^2 - \frac{3^3}{3} \right] - \left[3(2) + 2^2 - \frac{2^3}{3}$

$\displaystyle =1\frac{2}{3} m$

total distance travelled by the particle between $\displaystyle t = 3$ and $\displaystyle t = 4$

$\displaystyle =\int_{3}^{4} v dt$

$\displaystyle =\int_{3}^{4} (3 + 2t - t^2) dt$

$\displaystyle =\left[3t + t^2 - \frac{t^3}{3} \right]_{3}^{4}$

$\displaystyle =\left[3(4) + 4^2 - \frac{4^3}{3} \right] - \left[3(3) + 3^2 - \frac{3^3}{3} \right]$

$\displaystyle =6\frac{2}{3} - 9$

$\displaystyle =-2\frac{1}{3}$

$\displaystyle =2\frac{1}{3} m$

therefore,total distance travelled by the particle between $\displaystyle t=2$ and $\displaystyle t = 4$

$\displaystyle =1\frac{2}{3} + 2\frac{1}{3}$

$\displaystyle = 4m$

is that correct?

b)determine the maximum displacement of the particle from point $\displaystyle A$ before it reverses the direction of motion..

how can i find?