If the velocity of a particle is $$v = At + B{t^2},$$    where $$A$$ and $$B$$ are constants, then the distance travelled by it between $$1\,s$$  and $$2\,s$$  is

Solution :
Velocity of the particle is given as $$v = At + B{t^2}$$
where $$A$$ and $$B$$ are constants.
$$\eqalign{ & \Rightarrow \frac{{dx}}{{dt}} = At + B{t^2}\,\left[ {\because v = \frac{{dx}}{{dt}}} \right] \cr & \Rightarrow dx = \left( {At + B{t^2}} \right)dt \cr} $$
Integrating both sides, we get
$$\eqalign{ & \int_{{x_1}}^{{x_2}} d x = \int_1^2 {\left( {At + B{t^2}} \right)} \,dt \cr & \Rightarrow \Delta x = {x_2} - {x_1} = A\int_1^2 t dt + B\int_1^2 {{t^2}} dt \cr & = A\left[ {\frac{{{t^2}}}{2}} \right]_1^2 + B\left[ {\frac{{{t^3}}}{3}} \right]_1^2 \cr & = \frac{A}{2}\left( {{2^2} - {1^2}} \right) + \frac{B}{3}\left( {{2^3} - {1^3}} \right) \cr} $$
$$\therefore $$ Distance travelled between $$1\,s$$  and $$2\,s$$  is
$$\Delta x = \frac{A}{2} \times \left( 3 \right) + \frac{B}{3}\left( 7 \right) = \frac{{3A}}{2} + \frac{{7B}}{3}$$