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If \(f\) is a continuous function on \([a,b]\), then the average value \(f_{ave}\) of \(f\) over \([a,b]\) is given by \[\boxed{f_{ave}=\frac{1}{b-a}\int_a^b f(x)\,dx}\]

Break \([a,b]\) into \(n\) subintervals \([x_0,x_1],[x_1,x_2],\ldots,[x_{n-1},x_n]\) where \(x_i=a+i \Delta x\) and \(\Delta x=(b-a)/n\). Choose \(x_i^*\) from \([x_{i-1},x_i]\), \(i=1,2,\ldots,n\). The average value of \(f(x_1^*),f(x_2^*),\ldots,f(x_n^*)\) is \[\frac{f(x_1^*)+f(x_2^*)+\cdots +f(x_n^*)}{n} =\frac{\displaystyle\sum_{i=1}^n f(x_i^*)}{\frac{b-a}{\Delta x}} =\frac{1}{b-a}\sum_{i=1}^n f(x_i^*) \Delta x. \] This approximation of \(f_{ave}\) gets better as \(n\to \infty\). Thus \[f_{ave}=\lim_{n\to \infty}\frac{1}{b-a}\sum_{i=1}^n f(x_i^*) \Delta x =\frac{1}{b-a}\int_a^b f(x)\,dx. \]

The Mean Value Theorem for Integrals:
If \(f\) is a continuous function on \([a,b]\), then there is at least one number \(c\) in \([a,b]\) such that \(f(c)=f_{ave}\), i.e., \[f(c)=\frac{1}{b-a}\int_a^b f(x)\,dx.\]

Example. Find the average value \(f_{ave}\) of \(f(x)=x^2-2x+5\) over \([1,4]\) and find \(c\) in \([1,4]\) such that \(f(c)=f_{ave}\).
Solution. \[\begin{align*} f_{ave} &=\frac{1}{4-1}\int_1^4 (x^2-2x+5)\,dx\\ &=\frac{1}{3}\left. \left( \frac{x^3}{3}-x^2+5x \right) \right\vert_1^4\\ &=\frac{1}{3}\left[ \left( \frac{64}{3}-16+20 \right)- \left( \frac{1}{3}-1+5 \right) \right]\\ &=\frac{21}{3}\\ &=7. \end{align*}\] \[f(c)=f_{ave} \implies c^2-2c+5=7 \implies c^2-2c-2=0 \implies c=1\pm \sqrt{3}.\] Since \(1+\sqrt{3}\) is in \([1,4]\), \(c=1+\sqrt{3}\).

Average Value of a Function