91Ó°ÊÓ

A function, f(x), is said to be periodic if there exists a constant 'p' such that \(f(x+p) = f(x)\) for all x in the domain of the function. In other words, the graph of the function repeats itself after a certain interval, which is called the period of the function.

An antiderivative, F(x), of a function f(x) is a differentiable function whose derivative is equal to f(x). That is, if \(F'(x) = f(x)\), then F(x) is an antiderivative of f(x). Note that there can be multiple antiderivatives of a function, differing by a constant. For example, if F(x) is an antiderivative of f(x), then \(F(x) + C\) is also an antiderivative of f(x), where C is an arbitrary constant.

Let’s assume that F(x) is an antiderivative of the periodic function f(x) with a period of 'p'. So, \(F'(x) = f(x)\). Since f(x) is periodic, we know that \(f(x+p) = f(x)\). Now, we need to determine if F(x) is also periodic. To check if F(x) is periodic, we must check if \(F(x+p) = F(x) + C\) for some constant C. Note that only if C is equal to zero, the antiderivative is periodic. Using the fundamental theorem of calculus, we can analyze the difference between \(F(x+p)\) and \(F(x)\): $$ F(x+p) - F(x) = \int_x^{x+p} F'(t) dt = \int_x^{x+p} f(t) dt $$ Since f(t) is periodic with a period of 'p', the integral of f(t) over one period should always be the same, i.e., the integral value should be a constant. So, $$ \int_x^{x+p} f(t) dt = C $$ However, if the integral value C is different from zero, then the antiderivative F(x) is not periodic, as \(F(x+p) = F(x) + C\), where C ≠ 0. If C = 0, then F(x) is also periodic. Thus, we cannot assert that the antiderivative of a periodic function is always a periodic function. The antiderivative could be periodic only if the integral of the function over one period is equal to zero.