# Thread: Integral absolute minimum/maximum

1. ## Integral absolute minimum/maximum

Suppose f has absolute minimum value m and absolute maximum value M. Between what two values must the integral G lie?

I thought it was m(3) for abs min.

and M(3) for max. Is it 2 instead since 1-->3 is 2? Not sure. thanks for any help!

2. ## Re: Integral absolute minimum/maximum

What is a (Riemann) integral?

Let f be a function on [a,b] and $a=x_0 be a partition P of [a,b]. Let $L_P=\sum_{i=1}^nm_i(x_i-x_{i-1})$ where $m_i$ is the minimum (actually greatest lower bound) of f on $[x_{i-1},x_i]$. Similarly, $M_P=\sum_{i=1}^nM_i(x_i-x_{i-1})$ where $M_i$ is the maximum of f on $[x_{i-1},x_i]$.

The integral $G=\int_a^bf(x)dx$ is that unique number with $L_P\leq G\leq M_P$ for every partition P.

That's really the definition of integral. The great fundamental theorem of calculus allows "easy" computation of G.

In particular a partition with n=1, has $L_P=m(b-a)$ and $U_P=M(b-a)$ and so $m(b-a)\leq G \leq M(b-a)$

So for your specific problem $2m\leq G\leq2M$

3. ## Re: Integral absolute minimum/maximum

All that is correct, but you should probably just remember the general rule that if a function has a bound, $f(x)\le{M}$, then it's integral (if it exists) is bounded by M times the length of the interval, $\int_a^bf(x)\,dx\le{M(b-a)}$. If you draw a graph and compare the two areas, it should be clear.

- Hollywood

4. ## Re: Integral absolute minimum/maximum

Thank you both! I had a feeling you had to subtract 3 and 1 but I wasn't 100%. Both of your explanations really make it clear though!

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### value must be an integral

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