Concept preview

What you will be able to do

1. 01Formulate and compute the average value of a continuous function on $[a,\,b]$ using $f_{avg}=\dfrac{1}{b-a}\displaystyle\int_a^b f(x)\,dx$, and explain why the definite integral must be divided by the width $(b-a)$.
2. 02Interpret the average value geometrically as the height of the rectangle over $[a,\,b]$ that has the same area as the region under the curve.
3. 03State and apply the Mean Value Theorem for integrals: a continuous $f$ attains its own average, i.e. there is a $c$ in $[a,\,b]$ with $f(c)=f_{avg}$.

To average the numbers $4,\,6,\,8$ you compute $\dfrac{4+6+8}{3}=6$: a sum divided by the count. Now imagine the "numbers" are the heights of a curve $y=f(x)$ at every point of an interval $[a,\,b]$. There is no finite count to divide by, so we replace the sum with a definite integral, which is precisely the machine for adding a continuous sweep of values. The role once played by "divide by the count" is now played by "divide by the length of the interval," $(b-a)$. That gives the definition of the average value of a function: $$f_{avg}=\frac{1}{b-a}\int_a^b f(x)\,dx.$$