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I have the following question, and I am asked to prove it.

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What I understand is that $m$ and $M$ are lower and upper bounds that the integral from $[a,b]$ is contained in. What I don't get is that what is the $m(b-a)$. What exactly does that mean? I need to understand this context before I prove it. Is there a name for this inequality?

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    $m$ and $M$ are not lower and upper bounds for the integral; they're the bounds for the integrand, $f$ on the interval of integration. And $m(b-a)$ means "the product of the number $m$ with the number $b-a$, which is the width of the domain of integration. – 2017-01-10
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    $m(b-a)$ means $m$ multiplied by the quantity $(b-a)$ – 2017-01-10
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    Bolton, I know that. I mean what is the significance – 2017-01-10
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    There's a name for that inequality/theorem: monotonicity of the integral. – 2017-01-10
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    You can see $m(b-a)$ and $M(b-a)$ like rectangles of $\mathbb R^2$, this preposition tells you that if $f$ has got sup/inf, then there is a rectangle contained in the area of the graph of $f$. – 2017-01-10
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    Is this what your looking for? https://en.wikipedia.org/wiki/Mean_value_theorem#First_mean_value_theorem_for_definite_integrals – 2017-01-10
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    @Git Gud: monotonicity (or positivity) of the integral is the *reason* why this inequality is true, not its name. – 2017-01-10

3 Answers 3

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Rewritten as $$m\le\frac1{b-a}\int_a^b f(x)\,\mathrm d\mkern 1mu x\le M,$$ it is known as the Mean value inequality.

There are several interpretations of this inequality. One of them is this: if you drive from a city A to a city B, the distance between A and B is $d$ and the speed indicator in always between speed $v_\min$ and speed $v_\max$, then the time required to go from A to B will be between $v_\min\times d$ and $v_\max\times d$.

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In general, we have

$$\int_a^b\min_{t\in[a,b]}f(t)\ dx\le\int_a^bf(x)\ dx\le\int_a^b\max_{t\in[a,b]}f(t)\ dx$$

Here, we have $m<\min_{t\in[a,b]}f(t)$ and $M>\max_{t\in[a,b]}f(t)$, thus

$$\int_a^bm\ dx<\int_a^b\min_{t\in[a,b]}f(t)\ dx\le\int_a^bf(x)\ dx\le\int_a^b\max_{t\in[a,b]}f(t)\ dx<\int_a^bM\ dx$$

$$m(b-a)<\int_a^bf(x)\ dx

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$m \leq f(x) \Rightarrow \int_a^b mdx \leq \int_a^b f(x)dx \Rightarrow m \int_a^b 1dx \leq \int_a^b f(x)dx \Rightarrow m(b-a) \leq \int_a^b f(x)dx$.

Similarly, $\int_a^b f(x)dx \leq M(b-a)$.

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    I think proving the first implication is the point of the exercise. – 2017-01-10
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    i somehow doubt that, but to prove the first inequality you could suppose otherwise, i.e., that $\int_a^b mdx > \int_a^b f(x) dx$ and then we would have $m(b-a) > \int_a^b f(x) dx \Rightarrow m > \frac{1}{b-a} \int_a^b f(x)dx$ which is the average value of $f$ on the interval $[a,b]$. Since the average value is strictly less than $m$, there must exist a point $x$ at which $f(x) < m$, a contradiction. – 2017-01-10
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    The last sentence requires proof. There's no way to prove this without going to the definition of integral. – 2017-01-10