# College Math Teaching

## January 22, 2014

### Mean Value Theorem for integrals and it’s use in Taylor Polynomial approximations

Filed under: Uncategorized — collegemathteaching @ 3:12 am

First, what is the Mean Value Theorem for integrals? There are two common versions (calculus level versions; one can make these more general):

Version one: if $f$ is continuous over $[a,b]$ then there is some $c \in (a,b)$ such that $f(c)(b-a) = \int^b_a f(x) dx$. The proof is pretty easy: Let $F(t) = \int^t_af(x)dx$ then the Fundamental Theorem of Calculus says that $F$ is continuous on $[a,b]$ and differentiable on $(a,b)$ so the Mean Value Theorem applies: there is some $c \in (a,b)$ where $F'(c)=f(x) =\frac{F(b)-F(a)}{b-a} = \frac{1}{b-a} \int^b_a f(x)dx$ and the result follows.

This version has a nice geometric interpretation. If one interprets the definite integral as an area bounded by the graph of the function, the $x$ axis and the lines $x=a, x =b$ then the Mean Value Theorem says that there is a rectangle whose base is $b-a$ and whose height is $f(c)$ whose area is equal to the integral. But there is a second version:

Version Two: if $f$ is continuous over $[a,b]$ and $g$ is positive and continuous (actually, integrable and positive is enough) on $[a,b]$ then there is some $c \in (a,b)$ where $f(c)\int^b_a g(x) dx = \int^b_a f(x) g(x) dx$.

The proof of this result is not much harder than the first one. Since $f$ is continuous over $[a,b]$, $f$ attains both a maximum and a minimum value on the interval; say the maximum is $f(x_M)$ is the maximum; $f(x_m)$ is the minimum. So we have for all $x \in [a,b], f(x_m)g(x) \le f(x)g(x) \le f(x_M)g(x)$ Now integrate: $f(x_m)\int^b_ag(x)dx \le \int^b_af(x)g(x)dx \le f(x_M)\int^b_ag(x)dx$ Now divide through by $\int^b_a g(x) dx$ and note that the integral is positive. So $f(x_m) \le \frac{\int^b_af(x)g(x)dx}{\int^b_a g(x) dx} \le f(x_M)$. Now because $f$ is continuous, every value between $f(x_m), f(x_M)$ is attained by $f$ so there is at least one $c$ such that $f(c) = \frac{\int^b_af(x)g(x)dx}{\int^b_a g(x) dx}$ and the result follows.

So, what does this have to do with Taylor polynomials?

In a previous post we went over how to use integration by parts to obtain a Taylor polynomial for a function that has a sufficient number of derivatives. Here is the plan:

for our purposes, let $f$ have as many derivatives at $0$ as desired.

Now compute $\int^x_0f'(t)dt = f(x)-f(0)$. Now compute by using integration by parts: $u =f'(t), dv= dt, du =f"(t), v = t-x$ (if the assignment of $v$ seems strange, remember we can use ANY anti-derivative of $dv$.)

So we have $f'(t)(t-x)|^x_0 - \int^x_0 f"(t)(t-x)dt = f(x)-f(0)$ so by substitution we obtain $f(0)+f'(0)x - \int^x_0 f"(t)(t-x) dt =f(x)$

I’ll repeat the process so you can see what happens to the sign in front of the integral: we use parts again: $u =f"(t), dv = (t-x), du = f^{(3)}(t), v = \frac{1}{2}(t-x)^2$ and so we obtain (for the integral) $\frac{1}{2}f"(t)(t-x)^2|^x_0 -\frac{1}{2}\int^x_0 f^{(3)}(t)(t-x)^2dt =-\frac{1}{2}f"(0)x^2-\frac{1}{2}\int^x_0 f^{(3)}(t)(t-x)^2dt$ Now substitute this for the integral in the order 1 expression, and remember the negative sign: we obtain $f(0)+f'(0)x \frac{1}{2}f"(0)x^2+\frac{1}{2}\int^x_0 f^{(3)}(t)(t-x)^2dt =f(x)$

So we can proceed inductively; the important thing here is the remainder term after $n$ steps is $\pm \frac{1}{n!} \int^x_0 f^{(n+1)}(t)(t-x)^n dt$

A word on the sign: the negative occurs when $n$ is odd and positive when $n$ is even. So one can remove this ambiguity by replacing $(t-x)$ by $(x-t)$ and so the remainder formula becomes: $\frac{1}{n!} \int^x_0 f^{(n+1)}(t)(x-t)^n dt$.

Now this is still not the usual Lagrange or Cauchy remainder formula that many texts give. But we can get that from our Mean Value Theorem for integrals. Since the integrand is continuous over the interval $[0,x]$ the First Mean Value Theorem for Integrals says that there exists a $\zeta \in (0,x)$ where $\frac{1}{n!}f^{(n+1)}(\zeta)x(x-\zeta)^n$ is the remainder; that is the Cauchy form.

The Lagrange form comes from the Second Mean Value Theorem for Integrals: we know that there is a $\omega \in (0,x)$ where $\frac{1}{n!} \int^x_0 f^{(n+1)}(t)(x-t)^n dt = f^{(n+1)}(\omega)\frac{1}{n!}\int^x_0(x-t)^n dt = \frac{1}{(n+1)!}f^{(n+1)}(\omega)x^{(n+1)}$

That is the Lagrange version of the error term that one usually sees.

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