A bullet is fired from a gun. The force on the bullet is given by
$$F = 600 - 2 \times {10^5}t$$
where, $$F$$ is in newton and $$t$$ in second. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?

Solution :
To calculate impulse first of all calculate the time during which force becomes zero.
We have given, $$F = 600 - 2 \times {10^5}t$$
When, bullet leaves the barrel, the force on the bullet becomes zero.
$$\eqalign{ & {\text{So,}}\,600 - 2 \times {10^5}t = 0 \cr & \Rightarrow t = \frac{{600}}{{2 \times {{10}^5}}} \cr & = 3 \times {10^{ - 3}}s \cr} $$
Then, average impulse imparted to the bullet
$$\eqalign{ & I = \int_0^t F dt = \int_0^{3 \times {{10}^{ - 3}}} {\left( {600 - 2 \times {{10}^5}t} \right)} dt \cr & = \left[ {600t - \frac{{2 \times {{10}^5}{t^2}}}{2}} \right]_0^{3 \times {{10}^{ - 3}}} \cr & = 600 \times 3 \times {10^{ - 3}} - {10^5} \times {\left( {3 \times {{10}^{ - 3}}} \right)^2} \cr & = 1.8 - 0.9 = 0.9\;N - s \cr} $$
Alternative
As obtained in previous method, the time taken by bullet when it leaves the barrel
$$t = 3 \times {10^{ - 3}}\;s$$
Let $${F_1}$$ and $${F_2}$$ denote the forces at the time of firing of bullets i.e. at $$t = 0$$  and at the time of leaving the bullet i.e. at $$t = 3 \times {10^{ - 3}}\;s.$$
$$\eqalign{ & {F_1} = 600 - 2 \times {10^5} \times 0 = 600\;N \cr & {F_2} = 600 - 2 \times {10^5} \times 3 \times {10^{ - 3}} = 0 \cr} $$
Mean value of force $$F = \frac{1}{2}\left( {{F_1} + {F_2}} \right) = \frac{{600 + 0}}{2} = 300\;N$$
Thus, impulse $$ = F \times t = 300 \times 3 \times {10^{ - 3}} = 0.9\;N - s$$