Final answer:
The statement provided is false; the divergence of the integral of g(x) does not imply the divergence of the integral of f(x) if f(x) ≤ g(x). The convergence or divergence of an integral depends on the behavior of the function at infinity, which is not provided in this question.
Step-by-step explanation:
The statement "If f(x) ≤ g(x) and ∫ g(x) dx from 0 to ∞ diverges, then ∫ f(x) dx from 0 to ∞ also diverges" is false. To understand why let's consider the behavior of improper integrals, where the limits of integration extend to infinity. If f(x) ≤ g(x), it is possible for the integral of f(x) to converge even if the integral of g(x) diverges. For example, if f(x) approaches zero sufficiently fast, its integral from 0 to infinity could yield a finite value.
Let's consider the provided content. The graph of a function f(x) that is a horizontal line between 0 and 20 suggests that f(x) is constant in this interval and, therefore, its integral in this range is simply f(x) multiplied by the width of the interval. However, this does not directly relate to the behavior of the function or its integral as x approaches infinity. Since determining convergence or divergence of an integral depends on the exact function and its behavior at infinity, we cannot draw conclusions solely based on this limited interval.
Therefore, without additional information about the behavior of f(x) as x approaches infinity, we cannot state that ∫ f(x) dx from 0 to infinity diverges solely because ∫ g(x) dx does.