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## Average Value of a Function

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 $$[1,4]$$, $$c=1+\sqrt{3}$$.

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