Constant of Integration - Reason For A Constant Difference Between Antiderivatives

Reason For A Constant Difference Between Antiderivatives

This result can be formally stated in this manner: Let and be two everywhere differentiable functions. Suppose that for every real number x. Then there exists a real number C such that for every real number x.

To prove this, notice that . So F can be replaced by F-G and G by the constant function 0, making the goal to prove that an everywhere differentiable function whose derivative is always zero must be constant:

Choose a real number a, and let . For any x, the fundamental theorem of calculus says that

\begin{align} \int_a^x 0\,dt &= F(x)-F(a)\\ &= F(x)-C, \end{align}

which implies that . So F is a constant function.

Two facts are crucial in this proof. First, the real line is connected. If the real line were not connected, we would not always be able to integrate from our fixed a to any given x. For example, if we were to ask for functions defined on the union of intervals and, and if a were 0, then it would not be possible to integrate from 0 to 3, because the function is not defined between 1 and 2. Here there will be two constants, one for each connected component of the domain. In general, by replacing constants with locally constant functions, we can extend this theorem to disconnected domains. For example, there are two constants of integration for and infinitely many for so for example the general form for the integral of 1/x is:

\int {1 \over x}\,dx = \begin{cases}\ln \left|x \right| + C^- & x < 0\\ \ln \left|x \right| + C^+ & x > 0 \end{cases}

Second, F and G were assumed to be everywhere differentiable. If F and G are not differentiable at even one point, the theorem fails. As an example, let be the Heaviside step function, which is zero for negative values of x and one for non-negative values of x, and let . Then the derivative of F is zero where it is defined, and the derivative of G is always zero. Yet it's clear that F and G do not differ by a constant. Even if it is assumed that F and G are everywhere continuous and almost everywhere differentiable the theorem still fails. As an example, take F to be the Cantor function and again let G = 0.

Read more about this topic:  Constant Of Integration

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