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Fill in a table of implications for the integral of a function

Let f be a function such that the integral

    \[ A(x) = \int_a^x f(t) \, dt \]

exists for all x in an interval [a,b]. Consider the following statements:

a. f is continuous at c.
b. f is discontinuous at c.
c. f is increasing on (a,b).
d. f'(c) exists.
e. f' is continuous at c.

\alpha. A is continuous at c.
\beta. A is discontinuous at c.
\gamma. A is convex on (a,b).
\delta. A'(c) exists.
\varepsilon. A' is continuous at c.

Make a table to indicate by T whether the statement in row (a)-(e) implies the statement in column (\alpha) - (\varepsilon). Leave the cell in the table blank if there is no implication.

The requested table is as follows:

    \[ \begin{tabular}{c | c c c c c} & $\alpha$ & $\beta$ & $\gamma$ & $\delta$ & $\varepsilon$ \\ \hline a & T& & & T& \\ b & T& & & & \\ c & T& & T& & \\ d & T& & & T& \\ e & T& & & T&T  \end{tabular} \]

We know the first column has all T’s since the function A(x) = \int_a^x f(t) \, dt is continuous for any function f by the first fundamental theorem we know the derivative A'(x) exists at every point in the open interval (a,b). Since differentiability implies continuity (by example 7 on page 163 of Apostol), we then know that A(x) is continuous at any point c \in (a,b).

We know the second column can have no T’s anywhere since, by the same argument as above, the function A(x) is continuous at every point c \in (a,b). Thus, it cannot be discontinuous at any point c \in (a,b).

The statement a,b,d, and e cannot imply that A(x) is convex on (a,b) since they are statements regarding continuity and existence of derivatives. None of these properties have anything to do with convexity. Then, statement (c) does imply statement (\gamma) since we know that a function is convex if its derivative is increasing. Since f is the derivative of A and it is increasing, we have that A is convex.


  1. Anonymous says:

    I agree to thee all, but guys.
    Why b can’t go to delta? The Fundamental Theorem of Calculus says if ( f cont on p),then F'(p) exists and equals to f(p);
    it’s like proposition* implies #, but # doesnt require * to be true.
    I mean , behavior of F outside pf points where f has continuity, is out of the general control ,right?

      • Anonymous says:

        I know that for A to be convex in an interval its derivative needs to be increasing in that interval.
        But A'(x) = f(x) where f(x) is continuous but c does not mention anything about f being continuous, only that it is increasing.
        Therefore A'(x) may not be equal to f(x).
        I know the book has the solution that c implies gamma but I am struggling accepting that f is the derivative of A without stating that f is continuous.

  2. Anonymous says:

    Hi RoRi, shouldn’t d==>theta? With f'(c) exists, f(x) is continuous at c. Since A'(x)=f(x) ==> A'(x) is continuous at x=c.
    Thanks for the wonder effort by putting up the full solutions!

      • nu creation says:

        (a) and (d) don’t imply epsilon because all the FTC says is that if f is continuous at c then A'(c) = the number f(c). but A'(x) could be different from f(x) for any other x. then you couldn’t argue that the continuity of f at c implies the continuity of A’ at c because these are two different functions no matter how close you get to c. However, part (e) implies epsilon because the continuity of f’ at c implies the existence of a neighborhood around c in which f’ exists implying a neighborhood around c in which f is continuous implying a neighborhood around c in which A'(x) = f(x) for all x in the neighborhood. so there is a neighborhood around c such that the functions A’ and f are the same function so the continuity of f in this case implies the continuity of A’.

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