Application of calculus in kinematics- calculating net distance vs. total distance ??

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There is a set of questions where it says to find the net distance and the total distance from $t_1$ to $t_2$ when the function v(t) is given. I get that you obtain the displacement function s(t) by integrating v(t), but what exactly is the difference between the calculations for the net distance and the total distance?

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To find total distance traveled, you must break up the path into pieces based on where the velocity changes sign, and find displacement along each of those pieces. Otherwise, you get unwanted cancellation. So, if your $v(t)$ changes sign at $t_0, t_1, \dots,t_n$, then total distance traveled is $$D=\int_{t_0}^{t_1}|v(t)|\,dt+\int_{t_1}^{t_2}|v(t)|\,dt+\dots+\int_{t_{n-1}}^{t_n}|v(t)|\,dt,$$ whereas the net displacement blows that all away and just does $$d=\int_{t_0}^{t_n}v(t)\,dt.$$