With the Fundamental Theorem of Calculus discussed, this section begins a focus into integrals of functions with symmetry, application problems, a strategy for finding an average value of a function and then finish with a major result: the Mean Value Theorem for Integrals
Recall the concepts of symmetry about the \(x\)-axis and symmetry about the origin. These concepts lead to the labels of an even function and an odd function. These special properties can be exploited when calculating the definite integral. The following example shows what happens to an even function.
Consider the definite integral:
$$\int_{-3}^{3}f(x)dx=\color{red}{A_1}+\color{blue}{A_2}+\color{green}{A_3}+ \color{orange}{A_4}$$
Since \(f\) is an even function, \(\color{red}{A_1}=\color{orange}{A_4}\) and \(\color{blue}{A_2}=\color{green}{A_3}\).
This means the definite integral becomes:
$$\int_{-3}^{3}f(x)dx= \color{green}{A_3}+\color{green}{A_3}+\color{orange}{A_4}+\color{orange}{A_4}\implies 2\cdot\left(\color{green}{A_3}+\color{orange}{A_4}\right)$$
Notice that the blue and orange areas can be defined using another definite integral:
$$\color{green}{A_3}+\color{orange}{A_4}=\int_{0}^{3}f(x)dx$$
Therefore, because of the symmetry of \(f\):
$$\int_{-3}^{3}f(x)dx=2\int_{0}^{3}f(x)dx$$
This result also holds for \(2\int_{-3}^{0}f(x)dx\).
Let \(a\) be a positive real number and \(f\) an integrable function on the interval \([-a,a]\).
Using Riemann sums for illustration purposes, suppose a partition of \([a,b]\) is \(a,\;x_1,\;x_2,\;\ldots,\;x_{n-1},\;b\), with \(\Delta x=\large{\frac{b-a}{n}}\). Let \(f(x_k^*)\) denote the function value on the subinterval \([x_{k-1},\;x_k]\), then the average of these function values is: $$\frac{f(x_1^*)+f(x_2^*)+\ldots+f(x_n^*)}{n}$$ Since \(n=\large{\frac{b-a}{\Delta x}}\) , some algebra and summation notation gives: $$\frac{f(x_1^*)+f(x_2^*)+\ldots+f(x_n^*)}{(b-a)/\Delta x}=\frac{1}{b-a}\sum_{k=1}^nf(x_k^*)\Delta x$$ The symbol for average function value is \(\overline{f}\), and applying limits to the Riemann sum we get the following result:
The average value of an integrable function \(f\) on the interval \([a,b]\): $$\overline{f}=\frac{1}{b-a}\int_{a}^{b}f(x)dx$$
The following video shows an example on applying the average value of a function.
Recall from Chapter 4, the Mean Value Theorem for derivatives: If \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then there exists at least one \(c\) in \((a,b)\) where $$f^\prime(c)=\frac{f(b)-f(a)}{b-a}$$ meaning, there is at least one value where the slope of the tangent line equals the slope of the secant line. A similar result holds for integrals:
Let \(f\) be continuous on the interval \([a,b]\). There exists a point \(c\) in \((a,b)\) such that: $$f(c)=\overline{f}=\frac{1}{b-a}\int_{a}^{b}f(t)dt$$
In words, there is a value where the function equals the average value of the function.
The following video shows an example on applying the Mean Value Theorem.