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Chapter 2: Problem 9

If \(f\) is continuous on \([a, b], f(x) \geq 0\) on \([a, b]\) and \(\mathrm{f}\left(\mathrm{x}_{0}\right)>0\) for some \(\mathrm{x}_{0}\) in \([\mathrm{a}, \mathrm{b}]\), prove that \(\int_{a}^{b} f(x) d x>0 .\)

Short Answer

Expert verified
Question: Prove that the definite integral of a continuous function f(x) is greater than zero under the following conditions: 1. f(x) is continuous on the closed interval [a, b]. 2. f(x) is greater than or equal to zero on [a, b]. 3. There exists a point x_0 in [a, b] such that the value of f(x_0) is strictly greater than zero.

Step by step solution

01

Understand the properties of continuous functions

Since the function f(x) is continuous on the interval [a, b], we know that the function has no gaps, jumps, or sudden breaks in its graph within that interval. Moreover, there must be an interval around x_0 that has positive values.
02

Consider the function value at x_0

We are given that the function f(x) has a positive value at x_0, i.e., f(x_0) > 0. Since the function is continuous, there exists an interval around x_0, let's say [c, d], where c > a and d < b, such that the function f(x) is positive within this interval. Mathematically, f(x) > 0 for all x in [c, d].
03

Recall the relationship between definite integral and function values

The definite integral of a function can be thought of as the signed area between the function's graph and the x-axis. Since f(x) is always non-negative on [a, b], the definite integral will represent the non-negative area under the curve of f(x) on the interval [a, b]. Let's split this integral into three parts: \(\int_{a}^{b} f(x) d x = \int_{a}^{c} f(x) d x + \int_{c}^{d} f(x) d x + \int_{d}^{b} f(x) d x\)
04

Analyze the three parts of the integral

We will now analyze each part of the split integral: i) \(\int_{a}^{c} f(x) d x\): The function f(x) is non-negative in this interval. Therefore, the integral here will be greater than or equal to zero. (\(\int_{a}^{c} f(x) d x \geq 0\)) ii) \(\int_{c}^{d} f(x) d x\): The function f(x) is strictly positive in this interval, i.e, f(x) > 0. Therefore, the integral here will also be strictly positive. (\(\int_{c}^{d} f(x) d x > 0\)) iii) \(\int_{d}^{b} f(x) d x\): The function f(x) is non-negative in this interval. Therefore, the integral here will be greater than or equal to zero. (\(\int_{d}^{b} f(x) d x \geq 0\))
05

Combine the results of the three parts

Combining the results from Step 4, we have: \(\int_{a}^{b} f(x) d x = (\int_{a}^{c} f(x) d x) + (\int_{c}^{d} f(x) d x) + (\int_{d}^{b} f(x) d x) \geq 0 + \int_{c}^{d} f(x) d x\) Since \(\int_{c}^{d} f(x) d x > 0\), it follows that: \(\int_{a}^{b} f(x) d x > 0\) Hence, we have shown that the definite integral of f(x) on the interval [a, b] will always be greater than zero under the given conditions.

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Most popular questions from this chapter

Let \(\mathrm{f}\) be a function. Show that there is a parabola \(\mathrm{y}=\mathrm{Ax}^{2}+\mathrm{Bx}+\mathrm{C}\) that passes through the three points \((-\mathrm{h}, \mathrm{f}(-\mathrm{h})),(0, \mathrm{f}(0))\), and \((\mathrm{h}, \mathrm{f}(\mathrm{h}))\).

If \(\alpha\) and \(\phi\) are positive acute angles then prove that \(\phi<\int_{0}^{p} \frac{\mathrm{dx}}{\sqrt{\left(1-\sin ^{2} \alpha \sin ^{2} \mathrm{x}\right)}}<\frac{\varphi}{\sqrt{\left(1-\sin ^{2} \alpha \sin ^{2} \varphi\right)}} .\) If \(\alpha=\phi=1 / 6 \pi\), then prove that the integral lies between \(0.523\) and \(0.541\).

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the integrals as areas. Let f be an increasing function with \(\mathrm{f}(0)=0\), and assume that it has an elementary antiderivative. Then \(\mathrm{f}^{-1}\) is an increasing function, and \(\mathrm{f}^{-1}(0)=0\). Prove that if \(f^{-1}\) is elementary, then it also has an elementary antiderivative.
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