1. In solution (3), why do you need a definite integral in the integral region from 0 to 1 when 1 ≤ x < 2?

Always: [lower integral limit 0, upper integral limit x],

When the upper integral limit x is 1 ≤ x < 2, the lower integral limit is still 0! ! So f(x)=∫[ lower integral limit 0, upper integral limit x]f(t)dt.

Because the expressions of f(x) in 0 ≤ x < 1 and 1 ≤ x < 2 are different, the integration interval should be divided into two sections.

F(x)=∫[ lower integral limit 0, upper integral limit 1]f(t)dt+∫[ lower integral limit 1, upper integral limit x]f(t)dt.

2. In the solution (4), when X≥2, why do we need to do definite integration in the integration areas from 0 to 1 and 1 to 2?

When the upper integral limit X≥2, the lower integral limit is still 0! !

F(x)=∫[ lower integral limit 0, upper integral limit x]f(t)dt=∫[ lower integral limit 0, upper integral limit 1]f(t)dt+∫[ lower integral limit 1, upper integral limit 2] f (t) dt.

Is the sum of integrals in three intervals.