Since, $ F=\frac{\Delta p}{\Delta t} $ or $ \Delta p=F\Delta t $ we can say that momentum between 0 to 7 s isequal to the vector area enclosed by theforce-time graph from 0 to 7s. So, Change in linear momentum = vector area of triangle OAB+vectorarea ofsquare BCDE + vector area of triangle $ EFG+ $ vector area of square $ GHIJ+ $ vector area oftriangle JKL $ =\left[ \frac{1}{2}\times 1\times (-1) \right]+[2\times 2]+\left[ \frac{1}{2}\times 2\times (-2) \right] $ $ +\,[1\times 1]+\left[ \frac{1}{2}\times 1\times (-1) \right] $ $ =-\frac{1}{2}+4-2+1-\frac{1}{2}=2N-s $
