Q. 69 - Calculus

Question

Follow the given steps to give an alternative proof of the Mean Value Theorem for Integrals (Theorem 4.31): If $f$ is continuous on a closed interval $\left[a,b\right]$, then there exists some $c\in (a,b)$ such that:

$f\left(c\right)=\frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx$

(a) Define $F\left(x\right)={\int }_a^{x}f\left(t\right)dt$. What three things does the Second Fundamental Theorem of Calculus say about $F$?

(b) Why does the Mean Value Theorem apply to $F$ on $[a,b]$, and what conclusion can we obtain from the Mean Value Theorem?

(c) Show that, for the value $c$ that is guaranteed by the Mean Value Theorem, $f\left(c\right)=\frac{1}{b-a}{\int }_a^{b}f\left(t\right)dt$.

Step-by-Step Solution

Answer

Part (a) The three things that the Second Fundamental Theorem of Calculus say about $F$ are:

1. $F$ is an antiderivative of $f$, so $F\text{'}=f$.

2. $F$ is continuous on $\left[a,b\right]$.

3. $F$ is differentiable on $\left(a,b\right)$.

Part (b) The Mean Value Theorem states that if $f$ is any continuous function on $\left[a,b\right]$, then there is some $c\in \left[a,b\right]$ with $f\left(c\right)=\frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx$.

There exists some $c$ for $c\in \left(a,b\right)$ such that $F\text{'}\left(C\right)=\frac{F\left(b\right)-F\left(a\right)}{b-a}$.

Part (c) The value $c$ is guaranteed by the Mean Value Theorem.

Part (a) Step 1. Given information

We have to prove the Mean Value Theorem.

Part (a) Step 2. Define F x = ∫ a x f t d t . Tell the three things that the Second Fundamental Theorem of Calculus say about F .

The three things that the Second Fundamental Theorem of Calculus say about $F$ are:

1. $F$ is an antiderivative of $f$, so $F\text{'}=f$.

2. $F$ is continuous on $\left[a,b\right]$.

3. $F$ is differentiable on $\left(a,b\right)$.

Part (b) Step 1. Tell reason for why the Mean Value Theorem apply to F on a , b and also conclusion that we can get from the Mean Value Theorem.

The Mean Value Theorem states that if $f$ is any continuous function on $\left[a,b\right]$, then there is some $c\in \left[a,b\right]$ with $f\left(c\right)=\frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx$.

Hence, there exists some $c$ for $c\in \left(a,b\right)$ such that $F\text{'}\left(C\right)=\frac{F\left(b\right)-F\left(a\right)}{b-a}$.

Part (c) Step 1. Proof to show that the value c is guaranteed by the Mean Value Theorem.

$$\begin{gathered} f\left(c\right)=F\text{'}\left(c\right) \\ =\frac{F\left(b\right)-F\left(a\right)}{b-a} \\ =\frac{{\int }_a^{b}f\left(t\right)dt-{\int }_a^{a}f\left(t\right)dt}{b-a} \\ =\frac{{\int }_a^{b}f\left(t\right)dt-0}{b-a} \\ =\frac{{\int }_a^{b}f\left(t\right)dt}{b-a} \\ =\frac{1}{b-a}{\int }_a^{b}f\left(t\right)dt \end{gathered}$$

Therefore, the value $c$ is guaranteed by the Mean Value Theorem.