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### Appendix A.5 : Proof of Various Integral Properties

In this section we’ve got the proof of several of the properties we saw in the Integrals Chapter as well as a couple from the Applications of Integrals Chapter. #### Proof of : $$\int{{k\,f\left( x \right)\,dx}} = k\int{{f\left( x \right)\,dx}}$$ where $$k$$ is any number.

This is a very simple proof. Suppose that $$F\left( x \right)$$ is an anti-derivative of $$f\left( x \right)$$, i.e. $$F'\left( x \right) = f\left( x \right)$$. Then by the basic properties of derivatives we also have that,

${\left( {k\,F\left( x \right)} \right)^\prime } = k\,F'\left( x \right) = k\,f\left( x \right)$

and so $$k\,F\left( x \right)$$ is an anti-derivative of $$k\,f\left( x \right)$$, i.e. $${\left( {k\,F\left( x \right)} \right)^\prime } = k\,f\left( x \right)$$. In other words,

$\int{{k\,f\left( x \right)\,dx = k\,F\left( x \right) + c}} = k\int{{f\left( x \right)\,dx}}$ #### Proof of : $$\int{{f\left( x \right) \pm g\left( x \right)\,dx}} = \int{{f\left( x \right)\,dx}} \pm \int{{g\left( x \right)\,dx}}$$

This is also a very simple proof Suppose that $$F\left( x \right)$$ is an anti-derivative of $$f\left( x \right)$$ and that $$G\left( x \right)$$ is an anti-derivative of $$g\left( x \right)$$. So we have that $$F'\left( x \right) = f\left( x \right)$$ and $$G'\left( x \right) = g\left( x \right)$$. Basic properties of derivatives also tell us that

${\left( {F\left( x \right) \pm G\left( x \right)} \right)^\prime } = \,F'\left( x \right) \pm G'\left( x \right) = f\left( x \right) \pm g\left( x \right)$

and so $$\,F\left( x \right) + G\left( x \right)$$ is an anti-derivative of $$\,f\left( x \right) + g\left( x \right)$$ and $$\,F\left( x \right) - G\left( x \right)$$ is an anti-derivative of $$\,f\left( x \right) - g\left( x \right)$$. In other words,

$\int{{f\left( x \right) \pm g\left( x \right)\,dx}} = F\left( x \right) \pm G\left( x \right) + c = \int{{f\left( x \right)\,dx}} \pm \int{{g\left( x \right)\,dx}}$ #### Proof of : $$\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} = - \int_{{\,b}}^{{\,a}}{{f\left( x \right)\,dx}}$$

From the definition of the definite integral we have,

$\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} \hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\Delta x = \frac{{b - a}}{n}$

and we also have,

$\int_{{\,b}}^{{\,a}}{{f\left( x \right)\,dx}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} \hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\Delta x = \frac{{a - b}}{n}$

Therefore,

\begin{align*}\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} & = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\frac{{b - a}}{n}} \\ & = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\frac{{ - \left( {a - b} \right)}}{n}} \\ & = \mathop {\lim }\limits_{n \to \infty } \left( { - \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\frac{{a - b}}{n}} } \right)\\ & = - \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\frac{{a - b}}{n}} = - \int_{{\,b}}^{{\,a}}{{f\left( x \right)\,dx}}\end{align*} #### Proof of : $$\int_{{\,a}}^{{\,a}}{{f\left( x \right)\,dx}} = 0$$

From the definition of the definite integral we have,

\begin{align*}\int_{{\,a}}^{{\,a}}{{f\left( x \right)\,dx}} & = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} \hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\Delta x = \frac{{a - a}}{n} = 0\\ & = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\left( 0 \right)} \\ & = \mathop {\lim }\limits_{n \to \infty } 0\\ & = 0\end{align*} #### Proof of : $$\int_{{\,a}}^{{\,b}}{{cf\left( x \right)\,dx}} = c\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}}$$

From the definition of the definite integral we have,

\begin{align*}\int_{{\,a}}^{{\,b}}{{c\,f\left( x \right)\,dx}} & = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {c\,f\left( {x_i^*} \right)\Delta x} \\ & = \mathop {\lim }\limits_{n \to \infty } c\sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} \\ & = c\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} \\ & = c\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}}\end{align*}

Remember that we can pull constants out of summations and out of limits. #### Proof of : $$\int_{{\,a}}^{{\,b}}{{f\left( x \right) \pm g\left( x \right)\,dx}} = \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} \pm \int_{{\,a}}^{{\,b}}{{g\left( x \right)\,dx}}$$

First we’ll prove the formula for “+”. From the definition of the definite integral we have,

\begin{align*}\int_{{\,a}}^{{\,b}}{{f\left( x \right) + g\left( x \right)\,dx}} & = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left( {f\left( {x_i^*} \right) + \,g\left( {x_i^*} \right)} \right)\Delta x} \\ & = \mathop {\lim }\limits_{n \to \infty } \left( {\sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} + \sum\limits_{i = 1}^n {g\left( {x_i^*} \right)\Delta x} } \right)\\ & = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} + \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {g\left( {x_i^*} \right)\Delta x} \\ & = \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} + \int_{{\,a}}^{{\,b}}{{g\left( x \right)\,dx}}\end{align*}

To prove the formula for “-” we can either redo the above work with a minus sign instead of a plus sign or we can use the fact that we now know this is true with a plus and using the properties proved above as follows.

\begin{align*}\int_{{\,a}}^{{\,b}}{{f\left( x \right) - g\left( x \right)\,dx}} & = \int_{{\,a}}^{{\,b}}{{f\left( x \right) + \left( { - g\left( x \right)} \right)\,dx}}\\ & = \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} + \int_{{\,a}}^{{\,b}}{{\left( { - g\left( x \right)} \right)\,dx}}\\ & = \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} - \int_{{\,a}}^{{\,b}}{{g\left( x \right)\,dx}}\end{align*} #### Proof of : $$\int_{{\,a}}^{{\,b}}{{c\,dx}} = c\left( {b - a} \right)$$, $$c$$ is any number.

If we define $$f\left( x \right) = c$$ then from the definition of the definite integral we have,

\begin{align*}\int_{{\,a}}^{{\,b}}{{c\,dx}} & = \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}}\\ & = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} \hspace{0.25in}\hspace{0.25in}\Delta x = \frac{{b - a}}{n}\\ & = \mathop {\lim }\limits_{n \to \infty } \left( {\sum\limits_{i = 1}^n c } \right)\frac{{b - a}}{n}\\ & = \mathop {\lim }\limits_{n \to \infty } \left( {cn} \right)\frac{{b - a}}{n}\\ & = \mathop {\lim }\limits_{n \to \infty } c\left( {b - a} \right)\\ & = c\left( {b - a} \right)\end{align*} #### Proof of : If$$f\left( x \right) \ge 0$$ for $$a \le x \le b$$ then $$\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} \ge 0$$.

From the definition of the definite integral we have,

$\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} \hspace{0.25in}\hspace{0.25in}\Delta x = \frac{{b - a}}{n}$

Now, by assumption $$f\left( x \right) \ge 0$$ and we also have $$\Delta x > 0$$ and so we know that

$\sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} \ge 0$

So, from the basic properties of limits we then have,

$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} \ge \mathop {\lim }\limits_{n \to \infty } 0 = 0$

But the left side is exactly the definition of the integral and so we have,

$\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} \ge 0$ #### Proof of : If $$f\left( x \right) \ge g\left( x \right)$$ for$$a \le x \le b$$ then $$\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} \ge \int_{{\,a}}^{{\,b}}{{g\left( x \right)\,dx}}$$.

Since we have $$f\left( x \right) \ge g\left( x \right)$$ then we know that $$f\left( x \right) - g\left( x \right) \ge 0$$on $$a \le x \le b$$ and so by Property 8 proved above we know that,

$\int_{{\,a}}^{{\,b}}{{f\left( x \right) - g\left( x \right)\,dx}} \ge 0$

We also know from Property 4 that,

$\int_{{\,a}}^{{\,b}}{{f\left( x \right) - g\left( x \right)\,dx}} = \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} - \int_{{\,a}}^{{\,b}}{{g\left( x \right)\,dx}}$

So, we then have,

\begin{align*}\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} - \int_{{\,a}}^{{\,b}}{{g\left( x \right)\,dx}} & \ge 0\\ \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} & \ge \int_{{\,a}}^{{\,b}}{{g\left( x \right)\,dx}}\end{align*} #### Proof of : If $$m \le f\left( x \right) \le M$$ for $$a \le x \le b$$ then $$m\left( {b - a} \right) \le \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} \le M\left( {b - a} \right)$$.

Given $$m \le f\left( x \right) \le M$$ we can use Property 9 on each inequality to write,

$\int_{{\,a}}^{{\,b}}{{m\,dx}} \le \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} \le \int_{{\,a}}^{{\,b}}{{M\,dx}}$

Then by Property 7 on the left and right integral to get,

$m\left( {b - a} \right) \le \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} \le M\left( {b - a} \right)$ #### Proof of : $$\left| {\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}}} \right| \le \int_{{\,a}}^{{\,b}}{{\left| {f\left( x \right)\,} \right|dx}}$$

First let’s note that we can say the following about the function and the absolute value,

$- \left| {f\left( x \right)} \right| \le f\left( x \right) \le \left| {f\left( x \right)} \right|$

If we now use Property 9 on each inequality we get,

$\int_{{\,a}}^{{\,b}}{{ - \left| {f\left( x \right)} \right|\,dx}} \le \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} \le \int_{{\,a}}^{{\,b}}{{\left| {f\left( x \right)} \right|\,dx}}$

We know that we can factor the minus sign out of the left integral to get,

$- \int_{{\,a}}^{{\,b}}{{\left| {f\left( x \right)} \right|\,dx}} \le \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} \le \int_{{\,a}}^{{\,b}}{{\left| {f\left( x \right)} \right|\,dx}}$

Finally, recall that if $$\left| p \right| \le b$$ then $$- b \le p \le b$$ and of course this works in reverse as well so we then must have,

$\left| {\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}}} \right| \le \int_{{\,a}}^{{\,b}}{{\left| {f\left( x \right)\,} \right|dx}}$ #### Fundamental Theorem of Calculus, Part I

If $$f\left( x \right)$$ is continuous on $$\left[ {a,b} \right]$$ then,

$g\left( x \right) = \int_{{\,a}}^{{\,x}}{{f\left( t \right)\,\,dt}}$

is continuous on $$\left[ {a,b} \right]$$ and it is differentiable on $$\left( {a,b} \right)$$ and that,

$g'\left( x \right) = f\left( x \right)$

#### Proof

Suppose that $$x$$ and $$x + h$$ are in $$\left( {a,b} \right)$$. We then have,

$g\left( {x + h} \right) - g\left( x \right) = \int_{{\,a}}^{{\,x + h}}{{f\left( t \right)\,\,dt}} - \int_{{\,a}}^{{\,x}}{{f\left( t \right)\,\,dt}}$

Now, using Property 5 of the Integral Properties we can rewrite the first integral and then do a little simplification as follows.

\begin{align*}g\left( {x + h} \right) - g\left( x \right) = \left( {\int_{{\,a}}^{{\,x}}{{f\left( t \right)\,\,dt}} + \int_{{\,x}}^{{\,x + h}}{{f\left( t \right)\,\,dt}}} \right) - \int_{{\,a}}^{{\,x}}{{f\left( t \right)\,\,dt}}\\ & = \int_{{\,x}}^{{\,x + h}}{{f\left( t \right)\,\,dt}}\end{align*}

Finally assume that $$h \ne 0$$ and we get,

$\begin{equation}\frac{{g\left( {x + h} \right) - g\left( x \right)}}{h} = \frac{1}{h}\int_{{\,x}}^{{\,x + h}}{{f\left( t \right)\,\,dt}} \label{eq:eq1} \end{equation}$

Let’s now assume that $$h > 0$$ and since we are still assuming that $$x + h$$ are in $$\left( {a,b} \right)$$ we know that $$f\left( x \right)$$ is continuous on $$\left[ {x,x + h} \right]$$ and so by the Extreme Value Theorem we know that there are numbers $$c$$ and $$d$$ in $$\left[ {x,x + h} \right]$$ so that $$f\left( c \right) = m$$ is the absolute minimum of $$f\left( x \right)$$ in $$\left[ {x,x + h} \right]$$ and that $$f\left( d \right) = M$$ is the absolute maximum of $$f\left( x \right)$$ in $$\left[ {x,x + h} \right]$$.

So, by Property 10 of the Integral Properties we then know that we have,

$mh \le \int_{{\,x}}^{{\,x + h}}{{f\left( t \right)\,dt}} \le Mh$

Or,

$f\left( c \right)h \le \int_{{\,x}}^{{\,x + h}}{{f\left( t \right)\,dt}} \le f\left( d \right)h$

Now divide both sides of this by h to get,

$f\left( c \right) \le \frac{1}{h}\int_{{\,x}}^{{\,x + h}}{{f\left( t \right)\,dt}} \le f\left( d \right)$

and then use $$\eqref{eq:eq1}$$ to get,

$\begin{equation}f\left( c \right) \le \frac{{g\left( {x + h} \right) - g\left( x \right)}}{h} \le f\left( d \right) \label{eq:eq2} \end{equation}$

Next, if $$h < 0$$ we can go through the same argument above except we’ll be working on $$\left[ {x + h,x} \right]$$ to arrive at exactly the same inequality above. In other words, $$\eqref{eq:eq2}$$ is true provided $$h \ne 0$$.

Now, if we take $$h \to 0$$ we also have $$c \to x$$ and $$d \to x$$ because both $$c$$ and $$d$$ are between $$x$$ and $$x + h$$. This means that we have the following two limits.

$\mathop {\lim }\limits_{h \to 0} f\left( c \right) = \mathop {\lim }\limits_{c \to x} f\left( c \right) = f\left( x \right)\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\mathop {\lim }\limits_{h \to 0} f\left( d \right) = \mathop {\lim }\limits_{d \to x} f\left( d \right) = f\left( x \right)$

The Squeeze Theorem then tells us that,

$\begin{equation}\mathop {\lim }\limits_{h \to 0} \frac{{g\left( {x + h} \right) - g\left( x \right)}}{h} = f\left( x \right) \label{eq:eq3} \end{equation}$

but the left side of this is exactly the definition of the derivative of $$g\left( x \right)$$ and so we get that,

$g'\left( x \right) = f\left( x \right)$

So, we’ve shown that $$g\left( x \right)$$ is differentiable on $$\left( {a,b} \right)$$.

Now, the Theorem at the end of the Definition of the Derivative section tells us that $$g\left( x \right)$$ is also continuous on $$\left( {a,b} \right)$$. Finally, if we take $$x = a$$ or $$x = b$$ we can go through a similar argument we used to get $$\eqref{eq:eq3}$$ using one-sided limits to get the same result and so the theorem at the end of the Definition of the Derivative section will also tell us that $$g\left( x \right)$$ is continuous at $$x = a$$ or $$x = b$$ and so in fact $$g\left( x \right)$$ is also continuous on $$\left[ {a,b} \right]$$. #### Fundamental Theorem of Calculus, Part II

Suppose $$f\left( x \right)$$ is a continuous function on $$\left[ {a,b} \right]$$ and also suppose that $$F\left( x \right)$$ is any anti-derivative for $$f\left( x \right)$$. Then,

$\int_{{\,a}}^{{\,b}}{{f\left( x \right)dx}} = \left. {F\left( x \right)} \right|_a^b = F\left( b \right) - F\left( a \right)$

#### Proof

First let $$g\left( x \right) = \int_{{\,a}}^{{\,x}}{{f\left( t \right)\,\,dt}}$$ and then we know from Part I of the Fundamental Theorem of Calculus that $$g'\left( x \right) = f\left( x \right)$$ and so $$g\left( x \right)$$ is an anti-derivative of $$f\left( x \right)$$ on $$\left[ {a,b} \right]$$. Further suppose that $$F\left( x \right)$$ is any anti-derivative of $$f\left( x \right)$$ on $$\left[ {a,b} \right]$$ that we want to choose. So, this means that we must have,

$g'\left( x \right) = F'\left( x \right)$

Then, by Fact 2 in the Mean Value Theorem section we know that $$g\left( x \right)$$ and $$F\left( x \right)$$ can differ by no more than an additive constant on $$\left( {a,b} \right)$$. In other words, for $$a < x < b$$ we have,

$F\left( x \right) = g\left( x \right) + c$

Now because $$g\left( x \right)$$ and $$F\left( x \right)$$ are continuous on $$\left[ {a,b} \right]$$, if we take the limit of this as $$x \to {a^ + }$$ and $$x \to {b^ - }$$ we can see that this also holds if $$x = a$$ and $$x = b$$.

So, for $$a \le x \le b$$we know that $$F\left( x \right) = g\left( x \right) + c$$. Let’s use this and the definition of $$g\left( x \right)$$ to do the following.

\begin{align*}F\left( b \right) - F\left( a \right) & = \left( {g\left( b \right) + c} \right) - \left( {g\left( a \right) + c} \right)\\ & = g\left( b \right) - g\left( a \right)\\ & = \int_{{\,a}}^{{\,b}}{{f\left( t \right)\,\,dt}} + \int_{{\,a}}^{{\,a}}{{f\left( t \right)\,\,dt}}\\ & = \int_{{\,a}}^{{\,b}}{{f\left( t \right)\,\,dt}} + 0\\ & = \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,\,dx}}\end{align*}

Note that in the last step we used the fact that the variable used in the integral does not matter and so we could change the $$t$$’s to $$x$$’s. #### Average Function Value

The average value of a continuous function $$f\left( x \right)$$ over the interval $$\left[ {a,b} \right]$$ is given by,

${f_{avg}} = \frac{1}{{b - a}}\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}}$

#### Proof

We know that the average value of $$n$$ numbers is simply the sum of all the numbers divided by $$n$$ so let’s start off with that. Let’s take the interval $$\left[ {a,b} \right]$$ and divide it into $$n$$ subintervals each of length,

$\Delta x = \frac{{b - a}}{n}$

Now from each of these intervals choose the points $$x_1^*,x_2^*, \ldots ,x_n^*$$ and note that it doesn’t really matter how we choose each of these numbers as long as they come from the appropriate interval. We can then compute the average of the function values $$f\left( {x_1^*} \right),f\left( {x_2^*} \right), \ldots ,f\left( {x_n^*} \right)$$ by computing,

$\begin{equation}\frac{{f\left( {x_1^*} \right) + f\left( {x_2^*} \right) + \cdots + f\left( {x_n^*} \right)}}{n} \label{eq:eq4} \end{equation}$

Now, from our definition of $$\Delta x$$ we can get the following formula for $$n$$.

$n = \frac{{b - a}}{{\Delta x}}$

and we can plug this into $$\eqref{eq:eq4}$$ to get,

\begin{align*}\frac{{f\left( {x_1^*} \right) + f\left( {x_2^*} \right) + \cdots + f\left( {x_n^*} \right)}}{{\frac{{b - a}}{{\Delta x}}}} & = \frac{{\left[ {f\left( {x_1^*} \right) + f\left( {x_2^*} \right) + \cdots + f\left( {x_n^*} \right)} \right]\Delta x}}{{b - a}}\\ & = \frac{1}{{b - a}}\left[ {f\left( {x_1^*} \right)\Delta x + f\left( {x_2^*} \right)\Delta x + \cdots + f\left( {x_n^*} \right)\Delta x} \right]\\ & = \frac{1}{{b - a}}\sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} \end{align*}

Let’s now increase $$n$$. Doing this will mean that we’re taking the average of more and more function values in the interval and so the larger we chose $$n$$ the better this will approximate the average value of the function.

If we then take the limit as $$n$$ goes to infinity we should get the average function value. Or,

${f_{avg}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{b - a}}\sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} = \frac{1}{{b - a}}\,\,\,\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x}$

We can factor the $$\frac{1}{{b - a}}$$ out of the limit as we’ve done and now the limit of the sum should look familiar as that is the definition of the definite integral. So, putting in definite integral we get the formula that we were after.

${f_{avg}} = \frac{1}{{b - a}}\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}}$ #### The Mean Value Theorem for Integrals

If $$f\left( x \right)$$ is a continuous function on $$\left[ {a,b} \right]$$ then there is a number $$c$$ in $$\left[ {a,b} \right]$$ such that,

$\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} = f\left( c \right)\left( {b - a} \right)$

#### Proof

Let’s start off by defining,

$F\left( x \right) = \int_{{\,a}}^{{\,x}}{{f\left( t \right)\,dt}}$

Since $$f\left( x \right)$$ is continuous we know from the Fundamental Theorem of Calculus, Part I that $$F\left( x \right)$$ is continuous on $$\left[ {a,b} \right]$$, differentiable on $$\left( {a,b} \right)$$ and that $$F'\left( x \right) = f\left( x \right)$$.

Now, from the Mean Value Theorem we know that there is a number $$c$$ such that $$a < c < b$$ and that,

$F\left( b \right) - F\left( a \right) = F'\left( c \right)\left( {b - a} \right)$

However, we know that $$F'\left( c \right) = f\left( c \right)$$ and,

$F\left( b \right) = \int_{{\,a}}^{{\,b}}{{f\left( t \right)\,dt}} = \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}}\hspace{0.25in}\hspace{0.25in}F\left( a \right) = \int_{{\,a}}^{{\,a}}{{f\left( t \right)\,dt}} = 0$

So, we then have,

$\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} = f\left( c \right)\left( {b - a} \right)$ #### Work

The work done by the force $$F\left( x \right)$$ (assuming that $$F\left( x \right)$$ is continuous) over the range $$a \le x \le b$$ is,

$W = \int_{{\,a}}^{{\,b}}{{F\left( x \right)\,dx}}$

#### Proof

Let’s start off by dividing the range $$a \le x \le b$$ into $$n$$ subintervals of width $$\Delta x$$ and from each of these intervals choose the points $$x_1^*,x_2^*, \ldots ,x_n^*$$.

Now, if $$n$$ is large and because $$F\left( x \right)$$ is continuous we can assume that $$F\left( x \right)$$ won’t vary by much over each interval and so in the $$i$$th interval we can assume that the force is approximately constant with a value of $$F\left( x \right) \approx F\left( {x_i^*} \right)$$. The work on each interval is then approximately,

${W_i} \approx F\left( {x_i^*} \right)\Delta x$

The total work over $$a \le x \le b$$ is then approximately,

$W \approx \sum\limits_{i = 1}^n {{W_i}} = \sum\limits_{i = 0}^n {F\left( {x_i^*} \right)\Delta x}$

Finally, if we take the limit of this as $$n$$ goes to infinity we’ll get the exact work done. So,

$W = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 0}^n {F\left( {x_i^*} \right)\Delta x}$

This is, however, nothing more than the definition of the definite integral and so the work done by the force $$F\left( x \right)$$ over $$a \le x \le b$$ is,

$W = \int_{{\,a}}^{{\,b}}{{F\left( x \right)\,dx}}$