Yes.
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Possibly under certain conditions, but not generally. Consider a nonmeasurable set A, and define f(x) = 1 if x in A 0 otherwise. Then {1} is certainly measurable but the inverse image {x | f(x) = 1} = A is not measurable.
No.
zero
If the first derivative if a function is a constant that the original function has only one slope across its entire domain, so it is a line.
The indefinite integral is the anti-derivative - so the question is, "What function has this given function as a derivative". And if you add a constant to a function, the derivative of the function doesn't change. Thus, for example, if the derivative is y' = 2x, the original function might be y = x squared. However, any function of the form y = x squared + c (for any constant c) also has the SAME derivative (2x in this case). Therefore, to completely specify all possible solutions, this constant should be added.