The graph of the force as a function of time is in the figure below.

There are two portions of the graph

`F_x(t) = 1000 N "for" t in[0,2] ms` ``

`F_x(t) =-250*t +1500 "for" t in [2,6] ms`

To compute the total impulse we start from the impulse theorem: Force is the variation of impulse `P` in the time unit.

`F =(dP)/dt` , or equivalent `dP = F*dt`

Now we integrate to find the total impulse

`P = int_0^PdP =int_0^0.006F(t)dt =int_0^0.002 1000*dt +int_0.002^0.006 (1500-250*t)*dt=`

`=1000*t (0->0.002) +[1500*t -(250/2)*t^2] (0.002 ->0.006) =`

`=1000*0.002 +1500*0.006-1500*0.002-125*0.006^2+125*0.002^2 =`

`=2+9-3-0.0045+0.0005 =7.996 kg*m/s`

**Further Reading**

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