# Integral absolute minimum/maximum

• Feb 15th 2013, 08:19 PM
Steelers72
Integral absolute minimum/maximum
http://www.webassign.net/cgi-perl/sy...0f%28x%29dx%20

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!
• Feb 15th 2013, 09:47 PM
johng
Re: Integral absolute minimum/maximum
What is a (Riemann) integral?

Let f be a function on [a,b] and $\displaystyle a=x_0<x_1<...<x_n=b$ be a partition P of [a,b]. Let $\displaystyle L_P=\sum_{i=1}^nm_i(x_i-x_{i-1})$ where $\displaystyle m_i$ is the minimum (actually greatest lower bound) of f on $\displaystyle [x_{i-1},x_i]$. Similarly, $\displaystyle M_P=\sum_{i=1}^nM_i(x_i-x_{i-1})$ where $\displaystyle M_i$ is the maximum of f on $\displaystyle [x_{i-1},x_i]$.

The integral $\displaystyle G=\int_a^bf(x)dx$ is that unique number with $\displaystyle L_P\leq G\leq M_P$ 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 $\displaystyle L_P=m(b-a)$ and $\displaystyle U_P=M(b-a)$ and so $\displaystyle m(b-a)\leq G \leq M(b-a)$

So for your specific problem $\displaystyle 2m\leq G\leq2M$
• Feb 16th 2013, 09:08 AM
hollywood
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, $\displaystyle f(x)\le{M}$, then it's integral (if it exists) is bounded by M times the length of the interval, $\displaystyle \int_a^bf(x)\,dx\le{M(b-a)}$. If you draw a graph and compare the two areas, it should be clear.

- Hollywood
• Feb 16th 2013, 02:53 PM
Steelers72
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!