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Chapter 5: Problem 10

If \(f\) is continuous on \([a, b]\) and \(\int_{a}^{b}|f(x)| d x=0,\) what can you conclude about \(f ?\)

Short Answer

Expert verified
Answer: If the integral of the absolute value of a continuous function f over the interval [a, b] is equal to 0, we can conclude that the function f is identically zero over the interval [a, b].

Step by step solution

01

1. Recall the fundamental theorem of calculus

We'll start by recalling the fundamental theorem of calculus, which states that if a function f is continuous on [a, b], then its integral F on that interval exists and F'(x) = f(x) for all x in [a, b].
02

2. Understand what it means for the integral of the absolute value of f to be 0

The integral of the absolute value of f over the interval [a, b] is given by \(\int_{a}^{b}|f(x)| d x\). This quantifies the total accumulated signed area of the region enclosed by the function and the x-axis. Since the absolute value will always be non-negative, this means that if this integral is 0, there must be no area enclosed by the graph of |f(x)| and the x-axis.
03

3. Conclude that f must be zero over the interval [a, b]

Given that there is no area enclosed by |f(x)| and the x-axis, and since |f(x)| is non-negative, it must be the case that f(x) = 0 for all x in [a, b]. Thus, we can conclude that the function f is identically zero over the interval [a, b].

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Most popular questions from this chapter

Find the area of the region \(R\) bounded by the graph of \(f\) and the \(x\) -axis on the given interval. Graph \(f\) and show the region \(R\) $$f(x)=\left(1-x^{2}\right)^{-1 / 2} ;[-1 / 2, \sqrt{3} / 2]$$

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