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Could someone help me through this problem?

Prove the uniqueness of the integral. That is, show that $F'\equiv 0$ implies that F is a constant

Try to show the following properties of integrals:

1)$\displaystyle\int_{C} [f(z)+g(z)]\, dz=$$\displaystyle\int_{C}f(z)\, dz +\displaystyle\int_{C} g(z)\, dz$

2)$\displaystyle\int_{C} af(z)\, dz=a\displaystyle\int_{C} f(z)\, dz$

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Hint: Mean Value Theorem. –  Henning Makholm Apr 25 '12 at 0:56
    
I cannot believe you have no idea about how to be started on any of these three questions. By the way, questions 1 and 2 are unrelated to the unnumbered question. –  Did May 7 '12 at 6:43

1 Answer 1

up vote 1 down vote accepted

For any $a < b$, we can use the mean value theorem to conclude that there is a $x \in (a, b)$ so that $$ 0 = f'(x) = \frac{f(b) - f(a)}{b - a} $$ which implies $f(a) = f(b)$. Since this holds for any $a$ and $b$, we must have $f(x) = c$ for all $x$ for some constant $c$.

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