It should be noted that whoever exports 0 is a constant, that is, the value has nothing to do with it.

This problem is about the derivation of a to get 0, so it is: No matter what value A takes, (a) both = the same constant is set to B.

So when a takes 0,? (0) also = B. So? (1) =? (0)=B .

Evidence two,

∫ [A to A+T] FDX = ∫ [A to 0] FDX+∫ [0 to T] FDX+∫ [T to A+T] FDX

For the third integral substitution stage x=u+T, the Du Can of integral = ∫ [0 to a] f (u) is obtained.

Then cancel it with the first integral.