5 Integration 5.3 Riemann Sums 5.5 Substitution

5.4 The Fundamental Theorem of Calculus

In this section we will find connections between differential calculus (derivatives and antiderivatives) and integral calculus (definite integrals). These connections between the major ideas of calculus are important enough to be called the Fundamental Theorem of Calculus. These connections will also explain why we use the term indefinite integral for the set of all antiderivatives, and why we use such similar notations for antiderivatives and definite integrals.

Let $f(t)$ be a continuous function defined on $[a,b]$ . The definite integral $\int_{a}^{b}f(x)\ dx$ is the “area under $f\$ ” on $[a,b]$ . We can turn this concept into a function by letting the upper (or lower) bound vary.

Let $F(x)=\int_{a}^{x}f(t)\ dt$ . It computes the area under $f$ on $[a,x]$ as illustrated in Figure 5.4.1. We can study this function using our knowledge of the definite integral. For instance, $F(a)=0$ since $\int_{a}^{a}f(t)\ dt=0$ .

^†^†margin: Figure 5.4.1: The area of the shaded region is

F(x)=\int_{a}^{x}f(t)\ dt

The first part of the Fundamental Theorem of Calculus tells us how to find derivatives of these kinds of functions.

Theorem 5.4.1 The Fundamental Theorem of Calculus, Part 1

Let $f$ be continuous on $[a,b]$ and let $F(x)=\int_{a}^{x}f(t)\ dt$ . Then $F$ is a continuous differentiable function on $(a,b)$ , and

F\hskip 0.75pt^{\prime}(x)=f(x).

Proof

In order to see why this is true, we must compute $\displaystyle\lim_{h\to 0}\frac{F(x+h)-F(x)}{h}$ . Suppose $x$ and $x+h$ are in $[a,b]$ . Theorem 5.2.1 implies that

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

which we can rewrite as

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

This allows us to simplify the numerator of the difference quotient in our limit as follows:

$\displaystyle F(x+h)-F(x)$ $\displaystyle=\int_{a}^{x+h}f(t)\,dt-\int_{a}^{x}f(t)\,dt\quad\text{(by the % definition of $F$)}$

$\displaystyle=\int_{x}^{x+h}f(t)\,dt,$

so we see that

$\lim_{h\to 0}\frac{F(x+h)-F(x)}{h}=\lim_{h\to 0}\frac{1}{h}\int_{x}^{x+h}f(t)% \,dt.$

Assume for the moment that $h>0$ . Since $x$ and $x+h$ are both in $[a,b]$ and $f$ is continuous on $[a,b]$ , $f$ is also continuous on $[x,x+h]$ . Applying the Extreme Value Theorem (Theorem 3.1.1), we know that $f$ must have an absolute minimum value $f(u)=m$ and an absolute maximum value $f(v)=M$ on this interval. In other words, $m\leq f(t)\leq M$ whenever $x\leq t\leq x+h$ . Using Theorem 5.3.3, we can now say that

$\int_{x}^{x+h}m\,dt\leq\int_{x}^{x+h}f(t)\,dt\leq\int_{x}^{x+h}M\,dt.$

Computing the outer integrals, this becomes

$\displaystyle m(x+h-x)\leq\int_{x}^{x+h}f(t)$ $\displaystyle\,dt\leq M(x+h-x),\quad\text{or}$

$\displaystyle mh\leq\int_{x}^{x+h}f(t)$ $\displaystyle\,dt\leq Mh.$

Since $h>0$ , we may divide by $h$ to obtain

$f(u)=m\leq\frac{1}{h}\int_{x}^{x+h}f(t)\,dt\leq M=f(v).$

Now suppose that $h<0$ . Preceding as before, we know that $f$ has an absolute minimum value $f(u)=m$ and an absolute maximum value $f(v)=M$ on the interval $[x+h,x]$ . We know that $m\leq f(t)\leq M$ whenever $x+h\leq t\leq x$ , so we have

$\int_{x+h}^{x}m\,dt\leq\int_{x+h}^{x}f(t)\,dt\leq\int_{x+h}^{x}M\,dt.$

Once again we compute to obtain

$-mh\leq\int_{x+h}^{x}f(t)\,dt\leq-Mh.$

Since $-h>0$ , we can divide by $-h$ to obtain:

$\displaystyle m\leq-\frac{1}{h}\int_{x+h}^{x}f(t)$ $\displaystyle\,dt\leq M$

$\displaystyle f(u)=m\leq\frac{1}{h}\int_{x}^{x+h}f(t)$ $\displaystyle\,dt\leq M=f(v)\quad\text{(using \autoref{thm:defintprop}(2))}$

We are now ready to compute the desired limit,

$\lim_{h\to 0}\frac{F(x+h)-F(x)}{h}=\lim_{h\to 0}\frac{1}{h}\int_{x}^{x+h}f(t)% \,dt.$

Whether $h>0$ or $h<0$ , we know that

$f(u)\leq\frac{1}{h}\int_{x}^{x+h}f(t)\,dt\leq f(v),$

where $u$ and $v$ are both between $x$ and $x+h$ . Note that

$\lim_{h\to 0}(x+h)=x\text{\quad and \quad}\lim_{h\to 0}x=x,$

so the Squeeze Theorem (Theorem 1.3.5) says that

$\lim_{h\to 0}u=x\text{\quad and \quad}\lim_{h\to 0}v=x.$

Since $f$ is continuous at $x$ , we know that

$\lim_{h\to 0}f(u)=f(x)\text{\quad and \quad}\lim_{h\to 0}f(v)=f(x).$

Finally, we know that

$f(u)\leq\frac{1}{h}\int_{x}^{x+h}f(t)\,dt\leq f(v)\text{,}$

so applying the Squeeze Theorem again tells us that

$\lim_{h\to 0}\frac{1}{h}\int_{x}^{x+h}f(t)\,dt=f(x).$

Therefore $F\hskip 0.75pt^{\prime}(x)=f(x)$ as desired. Because the limit exists, Exercise 33 in Section 2.1 implies that $F$ is continuous as well, so that it is differentiable. ∎

Watch the video:
Fundamental Theorem of Calculus Part 1 from https://youtu.be/PGmVvIglZx8

Initially this seems simple, as demonstrated in the following example.

Example 5.4.1 Using the Fundamental Theorem of Calculus, Part 1

Let $\displaystyle F(x)=\int_{-5}^{x}(t^{2}+\sin t)\ dt$ . What is $F\hskip 0.75pt^{\prime}(x)$ ?

SolutionUsing the Fundamental Theorem of Calculus, we have
$F\hskip 0.75pt^{\prime}(x)=x^{2}+\sin x$ .

This simple example reveals something incredible: $F(x)$ is an antiderivative of $x^{2}+\sin x$ . Therefore, $F(x)=\frac{1}{3}x^{3}-\cos x+C$ for some value of $C$ . (We can find $C$ , but generally we do not care. We know that $F(-5)=0$ , which allows us to compute $C$ . In this case, $C=\cos(-5)+\frac{125}{3}$ .)

We have done more than found a complicated way of computing an antiderivative. Consider a function $f$ defined on an open interval containing $a$ , $b$ and $c$ . Suppose we want to compute $\int_{a}^{b}f(t)\ dt$ . First, let $F(x)=\int_{c}^{x}f(t)\ dt$ . Using the properties of the definite integral found in Theorem 5.2.1, we know

	$\displaystyle\int_{a}^{b}f(t)\ dt$	$\displaystyle=\int_{a}^{c}f(t)\ dt+\int_{c}^{b}f(t)\ dt$
		$\displaystyle=-\int_{c}^{a}f(t)\ dt+\int_{c}^{b}f(t)\ dt$
		$\displaystyle=-F(a)+F(b)$
		$\displaystyle=F(b)-F(a).$

We now see how indefinite integrals and definite integrals are related: we can evaluate a definite integral using antiderivatives. Furthermore, Theorem 5.1.1 told us that any other antiderivative $G$ differs from $F$ by a constant: $G(x)=F(x)+C$ . This means that $G(b)-G(a)=(F(b)+C)-(F(a)+C)=F(b)-F(a)$ , and the formula we’ve just found holds for any antiderivative. Consequently, it does not matter what value of $C$ we use, and we might as well let $C=0$ . This proves the second part of the Fundamental Theorem of Calculus.

Theorem 5.4.2 The Fundamental Theorem of Calculus, Part 2

Let $f$ be continuous on $[a,b]$ and let $F$ be any antiderivative of $f$ . Then

\int_{a}^{b}f(x)\ dx=F(b)-F(a).

Watch the video:
The Fundamental Theorem of Calculus. Part 2 from https://youtu.be/nHnZVFeQvNQ

Example 5.4.2 Using the Fundamental Theorem of Calculus, Part 2

We spent a great deal of time in the previous section studying $\int_{0}^{4}(4x-x^{2})\ dx$ . Using the Fundamental Theorem of Calculus, evaluate this definite integral.

SolutionWe need an antiderivative of $f(x)=4x-x^{2}$ . All antiderivatives of $f$ have the form $F(x)=2x^{2}-\frac{1}{3}x^{3}+C$ ; for simplicity, choose $C=0$ .

The Fundamental Theorem of Calculus states

\int_{0}^{4}(4x-x^{2})\ dx=F(4)-F(0)=\big{(}2(4)^{2}-\frac{1}{3}4^{3}\big{)}-% \big{(}0-0\big{)}=32-\frac{64}{3}=32/3.

This is the same answer we obtained using limits in the previous section, just with much less work.

Notation:

A special notation is often used in the process of evaluating definite integrals using the Fundamental Theorem of Calculus. Instead of explicitly writing $F(b)-F(a)$ , the notation $F(x)\Big{|}_{a}^{b}$ is used. Thus the solution to Example 5.4.2 would be written as:

\int_{0}^{4}(4x-x^{2})\ dx=\left.\left(2x^{2}-\frac{1}{3}x^{3}\right)\right|_{% 0}^{4}=\big{(}2(4)^{2}-\frac{1}{3}4^{3}\big{)}-\big{(}0-0\big{)}=32/3.

Example 5.4.3 Using the Fundamental Theorem of Calculus, Part 2

Evaluate the following definite integrals.

1.\ \int_{-2}^{2}x^{3}\ dx\quad 2.\ \int_{0}^{\pi}\sin x\ dx\qquad 3.\ \int_{0% }^{5}e^{t}\ dt\qquad 4.\ \int_{4}^{9}\sqrt{u}\ du\qquad 5.\ \int_{1}^{5}2\ dx

Solution

1.

$\displaystyle\int_{-2}^{2}x^{3}\ dx=\left.\frac{1}{4}x^{4}\right|_{-2}^{2}=% \left(\frac{1}{4}2^{4}\right)-\left(\frac{1}{4}(-2)^{4}\right)=0.$
2.

$\displaystyle\int_{0}^{\pi}\sin x\ dx=-\cos x\Big{|}_{0}^{\pi}=-\cos\pi-\big{(% }-\cos 0\big{)}=1+1=2.$
(This is interesting; it says that the area under one “hump” of a sine curve is 2.)
3.

$\displaystyle\int_{0}^{5}e^{t}\ dt=e^{t}\Big{|}_{0}^{5}=e^{5}-e^{0}=e^{5}-1% \approx 147.41.$
4.

$\displaystyle\int_{4}^{9}\sqrt{u}\ du=\int_{4}^{9}u^{\frac{1}{2}}\ du=\left.% \frac{2}{3}u^{\frac{3}{2}}\right|_{4}^{9}=\frac{2}{3}\left(9^{\frac{3}{2}}-4^{% \frac{3}{2}}\right)=\frac{2}{3}\big{(}27-8\big{)}=\frac{38}{3}.$
5.

$\displaystyle\int_{1}^{5}2\ dx=2x\Big{|}_{1}^{5}=2(5)-2=2(5-1)=8.$

This integral is interesting; the integrand is a constant function, hence we are finding the area of a rectangle with width $(5-1)=4$ and height 2. Notice how the evaluation of the definite integral led to $2(4)=8$ .

In general, if $c$ is a constant, then $\int_{a}^{b}c\ dx=c(b-a)$ .

The Fundamental Theorem of Calculus and the Chain Rule

Part 1 of the Fundamental Theorem of Calculus (FTC) states that given $\displaystyle F(x)=\int_{a}^{x}f(t)\ dt$ , $F\hskip 0.75pt^{\prime}(x)=f(x)$ . Using other notation, $\displaystyle\frac{d}{dx}\big{(}F(x)\big{)}=f(x)$ . While we have just practiced evaluating definite integrals, sometimes finding antiderivatives is impossible and we need to rely on other techniques to approximate the value of a definite integral. Functions written as $F(x)=\int_{a}^{x}f(t)\ dt$ are useful in such situations.

It may be of further use to compose such a function with another. As an example, we may compose $F(x)$ with $g(x)$ to get

F\big{(}g(x)\big{)}=\int_{a}^{g(x)}f(t)\ dt.

What is the derivative of such a function? The Chain Rule can be employed to state

\frac{d}{dx}\Big{(}F\big{(}g(x)\big{)}\Big{)}=F\hskip 0.75pt^{\prime}\big{(}g(% x)\big{)}g\hskip 0.75pt^{\prime}(x)=f\big{(}g(x)\big{)}g\hskip 0.75pt^{\prime}% (x).

An example will help us understand this.

Example 5.4.4 The FTC, Part 1, and the Chain Rule

Find the derivative of $\displaystyle F(x)=\int_{2}^{x^{2}}\ln t\ dt$ .

SolutionWe can view $F(x)$ as being the function $\displaystyle G(x)=\int_{2}^{x}\ln t\ dt$ composed with $g(x)=x^{2}$ ; that is, $F(x)=G\big{(}g(x)\big{)}$ . The Fundamental Theorem of Calculus states that $G\hskip 0.75pt^{\prime}(x)=\ln x$ . The Chain Rule gives us

	$\displaystyle F\hskip 0.75pt^{\prime}(x)$	$\displaystyle=G\hskip 0.75pt^{\prime}\big{(}g(x)\big{)}g\hskip 0.75pt^{\prime}% (x)$
		$\displaystyle=\ln(g(x))g\hskip 0.75pt^{\prime}(x)$
		$\displaystyle=\ln(x^{2})2x$
		$\displaystyle=2x\ln x^{2}$

Normally, the steps defining $G(x)$ and $g(x)$ are skipped.

Practice this once more.

Example 5.4.5 The FTC, Part 1, and the Chain Rule

Find the derivative of $\displaystyle F(x)=\int_{\cos x}^{5}t^{3}\ dt.$

SolutionNote that $\displaystyle F(x)=-\int_{5}^{\cos x}t^{3}\ dt$ . Viewed this way, the derivative of $F$ is straightforward:

F\hskip 0.75pt^{\prime}(x)=\sin x\cos^{3}x.

Understanding Motion with the Fundamental Theorem of Calculus

We established, starting with Key Idea 2.2.1, that the derivative of a position function is a velocity function, and the derivative of a velocity function is an acceleration function. Now consider definite integrals of velocity and acceleration functions. Specifically, if $v(t)$ is a velocity function, what does $\displaystyle\int_{a}^{b}v(t)\ dt$ mean?

The Fundamental Theorem of Calculus states that

\int_{a}^{b}v(t)\ dt=V(b)-V(a),

where $V(t)$ is any antiderivative of $v(t)$ . Since $v(t)$ is a velocity function, $V(t)$ must be a position function, and $V(b)-V(a)$ measures a change in position, or displacement.

How would we measure total distance traveled? We have to consider the intervals when $v(t)\geq 0$ and when $v(t)\leq 0$ . Therefore,

\text{total distance traveled}=\int_{a}^{b}\left\lvert v(t)\right\rvert\ dt.

Example 5.4.6 Finding displacement and total distance traveled

A ball is thrown straight up with velocity given by $v(t)=-32t+20$ ft/s, where $t$ is measured in seconds. Find, and interpret,

1.\ \int_{0}^{1}v(t)\ dt\qquad\text{and}\qquad 2.\ \int_{0}^{1}\left\lvert v(t% )\right\rvert\ dt.

Solution

1.

Using the Fundamental Theorem of Calculus, we have

$\displaystyle\int_{0}^{1}v(t)\ dt$ $\displaystyle=\int_{0}^{1}(-32t+20)\ dt$

$\displaystyle=-16t^{2}+20t\Big{|}_{0}^{1}$

$\displaystyle=4\text{ ft}.$

Thus if a ball is thrown straight up into the air with velocity $v(t)=-32t+20$ , the height of the ball, 1 second later, will be 4 feet above the initial height. We will see in part 2. that the distance traveled is much farther. It has gone up to its peak and is falling down, but the difference between its height at $t=0$ and $t=1$ is 4 ft.
2.

Here we are trying to find the total distance traveled by the ball. We must first consider where $v(t)>0$ and $v(t)<0$ .

$\displaystyle v(t)=-32t+20$ $\displaystyle=0$

$\displaystyle-32t$ $\displaystyle=-20$

$\displaystyle t$ $\displaystyle=\frac{5}{8}$

This means $v(t)>0$ for $t<\frac{5}{8}$ and $v(t)<0$ for $t>\frac{5}{8}$ so we have

$\displaystyle\int_{0}^{1}\left\lvert v(t)\right\rvert\ dt$ $\displaystyle=\int_{0}^{5/8}v(t)\ dt+\int_{5/8}^{1}-v(t)\ dt$

$\displaystyle=\int_{0}^{5/8}-32t+20\ dt+\int_{5/8}^{1}32t-20\ dt$

$\displaystyle=\frac{34}{4}=8.5\text{ ft}.$

Integrating a rate of change function gives total change. Velocity is the rate of position change; integrating velocity gives the total change of position, i.e., displacement.

Integrating a speed function gives a similar, though different, result. Speed is also the rate of position change, but does not account for direction. So integrating a speed function gives total change of position, without the possibility of “negative position change.” Hence the integral of a speed function gives distance traveled.

As acceleration is the rate of velocity change, integrating an acceleration function gives total change in velocity. We do not have a simple term for this analogous to displacement. If $a(t)=5$ miles/h ${}^{2}$ and $t$ is measured in hours, then

\int_{0}^{3}a(t)\ dt=15

means the velocity has increased by 15m/h from $t=0$ to $t=3$ .

The Mean Value Theorem and Average Value

^†^†margin: (a) (b) (c) (d) Figure 5.4.2: A graph of a function

f

to introduce the Mean Value Theorem and differently sized rectangles giving upper and lower bounds on

\int_{1}^{4}f(x)\ dx

; the last rectangle matches the area exactly.

Consider the graph of a function $f$ in Figure 5.4.2(a) and the area defined by $\int_{1}^{4}f(x)\ dx$ . Three rectangles are then drawn; in (b), the height of the rectangle is greater than $f$ on $[1,4]$ , hence the area of this rectangle is is greater than $\int_{1}^{4}f(x)\ dx$ .

In (c), the height of the rectangle is smaller than $f$ on $[1,4]$ , hence the area of this rectangle is less than $\int_{1}^{4}f(x)\ dx$ .

Finally, in (d) the height of the rectangle is such that the area of the rectangle is exactly that of $\int_{1}^{4}f(x)\ dx$ . Since rectangles that are “too big ”, as in (b), and rectangles that are “too little,” as in (c), give areas greater/lesser than $\int_{1}^{4}f(x)\ dx$ , it makes sense that there is a rectangle, whose top intersects $f(x)$ somewhere on $[1,4]$ , whose area is exactly that of the definite integral.

We state this idea formally in a theorem.

Theorem 5.4.3 The Mean Value Theorem of Integration

Let $f$ be continuous on $[a,b]$ . There exists a value $c$ in $(a,b)$ such that

\int_{a}^{b}f(x)\ dx=f(c)(b-a).

This is an existential statement; $c$ exists, but we do not provide a method of finding it. Theorem 5.4.3 is directly connected to the Mean Value Theorem of Differentiation, given as Theorem 3.2.1.

Proof

If $a=b$ , then $\int_{a}^{a}f(x)\,dx=0=f(a)(a-a)$ . Otherwise, we define the following for $x$ in $[a,b]$ :

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

Applying Theorem 5.4.1 we know that $F$ is differentiable on $(a,b)$ and that $F\hskip 0.75pt^{\prime}(x)=f(x)$ for any $x$ in $(a,b)$ . We may now apply the Mean Value Theorem for Differentiation (Theorem 3.2.1) to see that there is a value $c$ in $(a,b)$ such that

$F^{\prime}(c)=\frac{F(b)-F(a)}{b-a}.$

Note that $F^{\prime}(c)=f(c)$ and that $F(b)-F(a)=\int_{a}^{b}f(x)\,dx$ by Theorem 5.4.2. Therefore we can rewrite our equation as:

$\displaystyle f(c)$ $\displaystyle=\frac{\int_{a}^{b}f(x)\,dx}{b-a},\text{ or}$

$\displaystyle f(c)(b-a)$ $\displaystyle=\int_{a}^{b}f(x)\,dx.\qed$

We demonstrate the principles involved in this version of the Mean Value Theorem in the following example.

^†^†margin: Figure 5.4.3: A graph of

y=\sin x

[0,\pi]

and the rectangle guaranteed by the Mean Value Theorem.

Example 5.4.7 Using the Mean Value Theorem

Consider $\displaystyle\int_{0}^{\pi}\sin x\ dx$ . Find a value $c$ guaranteed by the Mean Value Theorem.

SolutionWe first need to evaluate $\displaystyle\int_{0}^{\pi}\sin x\ dx$ . (This was previously done in Example 5.4.3.)

\int_{0}^{\pi}\sin x\ dx=-\cos x\Big{|}_{0}^{\pi}=2.

Thus we seek a value $c$ in $[0,\pi]$ such that $\pi\sin c=2$ .

\pi\sin c=2\quad\Rightarrow\quad\sin c=\frac{2}{\pi}\quad\Rightarrow\quad c=% \sin^{-1}\left(\frac{2}{\pi}\right)\approx 0.69.

In Figure 5.4.3 $\sin x$ is sketched along with a rectangle with height $\sin(0.69)$ . The area of the rectangle is the same as the area under $\sin x$ on $[0,\pi]$ .

^†^†margin: Figure 5.4.4: On top, a graph of

y=f(x)

and the rectangle guaranteed by the Mean Value Theorem. Below,

y=f(x)

is shifted down by

f(c)

; the resulting “area under the curve” is 0.

Let $f$ be a function on $[a,b]$ with $c$ such that $f(c)(b-a)=\int_{a}^{b}f(x)\ dx$ . Consider $\int_{a}^{b}\big{(}f(x)-f(c)\big{)}\ dx$ :

	$\displaystyle\int_{a}^{b}\big{(}f(x)-f(c)\big{)}\ dx$	$\displaystyle=\int_{a}^{b}f(x)-\int_{a}^{b}f(c)\ dx$
		$\displaystyle=f(c)(b-a)-f(c)(b-a)$
		$\displaystyle=0.$

When $f(x)$ is shifted by $-f(c)$ , the amount of area under $f$ above the $x$ -axis on $[a,b]$ is the same as the amount of area below the $x$ -axis above $f$ ; see Figure 5.4.4 for an illustration of this. In this sense, we can say that $f(c)$ is the average value of $f$ on $[a,b]$ .

The value $f(c)$ is the average value in another sense. First, recognize that the Mean Value Theorem can be rewritten as

f(c)=\frac{1}{b-a}\int_{a}^{b}f(x)\ dx,

for some value of $c$ in $[a,b]$ . Next, partition the interval $[a,b]$ into $n$ equally spaced subintervals, $a=x_{1}<x_{2}<\dots<x_{n+1}=b$ and choose any $c_{i}$ in $[x_{i},x_{i+1}]$ . The average of the numbers $f(c_{1})$ , $f(c_{2})$ , …, $f(c_{n})$ is:

\frac{1}{n}\Big{(}f(c_{1})+f(c_{2})+\dots+f(c_{n})\Big{)}=\frac{1}{n}\sum_{i=1% }^{n}f(c_{i}).

Multiply this last expression by 1 in the form of $\frac{(b-a)}{(b-a)}$ :

	$\displaystyle\frac{1}{n}\sum_{i=1}^{n}f(c_{i})$	$\displaystyle=\sum_{i=1}^{n}f(c_{i})\frac{1}{n}$
		$\displaystyle=\sum_{i=1}^{n}f(c_{i})\frac{1}{n}\frac{(b-a)}{(b-a)}$
		$\displaystyle=\frac{1}{b-a}\sum_{i=1}^{n}f(c_{i})\frac{b-a}{n}$
		$\displaystyle=\frac{1}{b-a}\sum_{i=1}^{n}f(c_{i})\Delta x\quad\text{% \scriptsize(where $\Delta x=(b-a)/n$)}$

Now take the limit as $n\to\infty$ :

\lim_{n\to\infty}\frac{1}{b-a}\sum_{i=1}^{n}f(c_{i})\Delta x\quad=\quad\frac{1% }{b-a}\int_{a}^{b}f(x)\ dx\quad=\quad f(c).

This tells us this: when we evaluate $f$ at $n$ (somewhat) equally spaced points in $[a,b]$ , the average value of these samples is $f(c)$ as $n\to\infty$ .

This leads us to a definition.

Definition 5.4.1 The Average Value of $f$ on $[a,b]$

Let $f$ be continuous on $[a,b]$ . The average value of $\mathbf{f}$ on $\mathbf{[a,b]}$ is $f(c)$ , where $c$ is a value in $[a,b]$ guaranteed by the Mean Value Theorem. I.e.,

\text{Average Value of $f$ on $[a,b]$}=\frac{1}{b-a}\int_{a}^{b}f(x)\ dx.

An application of this definition is given in the following example.

Example 5.4.8 Finding the average value of a function

An object moves back and forth along a straight line with a velocity given by $v(t)=(t-1)^{2}$ on $[0,3]$ , where $t$ is measured in seconds and $v(t)$ is measured in ft/s.

What is the average velocity of the object?

SolutionBy our definition, the average velocity is:

\frac{1}{3-0}\int_{0}^{3}(t-1)^{2}\ dt=\frac{1}{3}\int_{0}^{3}\big{(}t^{2}-2t+% 1\big{)}\ dt=\left.\frac{1}{3}\left(\frac{1}{3}t^{3}-t^{2}+t\right)\right|_{0}% ^{3}=1\text{ ft/s}.

We can understand the above example through a simpler situation. Suppose you drove 100 miles in 2 hours. What was your average speed? The answer is simple: displacement/time = 100 miles/2 hours = 50 mph.

What was the displacement of the object in Example 5.4.8? We calculate this by integrating its velocity function: $\int_{0}^{3}(t-1)^{2}\ dt=3$ ft. Its final position was 3 feet from its initial position after 3 seconds: its average velocity was 1 ft/s.

This section has laid the groundwork for a lot of great mathematics to follow. The most important lesson is this: definite integrals can be evaluated using antiderivatives. Since the previous section established that definite integrals are the limit of Riemann sums, we can later create Riemann sums to approximate values other than “area under the curve,” convert the sums to definite integrals, then evaluate these using the Fundamental Theorem of Calculus. This will allow us to compute the work done by a variable force, the volume of certain solids, the arc length of curves, and more.

The downside is this: generally speaking, computing antiderivatives is much more difficult than computing derivatives. Much of our time in Calculus II will be devoted to techniques of finding antiderivatives so that a wide variety of definite integrals can be evaluated.

Exercises 5.4

Terms and Concepts

1.

How are definite and indefinite integrals related?
2.

What constant of integration is most commonly used when evaluating definite integrals?
3.

T/F: If $f$ is a continuous function, then $\displaystyle F(x)=\int_{a}^{x}f(t)\ dt$ is also a continuous function.
4.

The definite integral can be used to find “the area under a curve.” Give two other uses for definite integrals.

Problems

In Exercises 5–34, use the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral.

5.

$\displaystyle\int_{1}^{3}(3x^{2}-2x+1)\ dx$
6.

$\displaystyle\int_{0}^{4}(x-1)^{2}\ dx$
7.

$\displaystyle\int_{-1}^{1}(x^{3}-x^{5})\ dx$
8.

$\displaystyle\int_{\pi/2}^{\pi}\cos x\ dx$
9.

$\displaystyle\int_{0}^{\pi/4}\sec^{2}x\ dx$
10.

$\displaystyle\int_{1}^{e}\frac{1}{x}\ dx$
11.

$\displaystyle\int_{-2}^{-1}(4-2x^{3})\ dx$
12.

$\displaystyle\int_{0}^{\pi}(2\cos x-2\sin x)\ dx$
13.

$\displaystyle\int_{1}^{3}e^{x}\ dx$
14.

$\displaystyle\int_{0}^{4}\sqrt{t}\ dt$
15.

$\displaystyle\int_{9}^{25}\frac{1}{\sqrt{t}}\ dt$
16.

$\displaystyle\int_{1}^{8}\sqrt[3]{x}\ dx$
17.

$\displaystyle\int_{1}^{2}\frac{1}{x}\ dx$
18.

$\displaystyle\int_{1}^{2}\frac{1}{x^{2}}\ dx$
19.

$\displaystyle\int_{0}^{1}x^{3}\ dx$
20.

$\displaystyle\int_{0}^{1}x^{100}\ dx$
21.

$\displaystyle\int_{-10}^{-5}3\ dx$
22.

$\displaystyle\int_{\pi/6}^{\pi/3}\csc x\cot x\ dx$
23.

$\displaystyle\int_{0}^{2}\left\lvert x^{2}-1\right\rvert\ dx$
24.

$\displaystyle\int_{0}^{3}\left\lvert 1-2x\right\rvert\ dx$
25.

$\displaystyle\int_{-1}^{2}(u+4)(2u+1)\ du$
26.

$\displaystyle\int_{1}^{9}\frac{1+\sqrt{x}+x}{\sqrt{x}}\ dx$
27.

$\displaystyle\int_{\pi/7}^{\pi}\sin^{2}x+\cos^{2}x\ dx$
28.

$\displaystyle\int_{-\pi/4}^{\pi/4}2+\tan^{2}\theta\ d\theta$
29.

$\displaystyle\int_{0}^{\pi/4}\sec t(\sec t+\tan t)\ dt$
30.

$\displaystyle\int_{\pi/6}^{\pi/2}\frac{\sin 2x}{\sin x}\ dx$
31.

$\displaystyle\int_{1}^{4}\frac{4+6u}{\sqrt{u}}\ du$
32.

$\displaystyle\int_{0}^{\pi/3}\frac{\sin\theta+\sin\theta\tan^{2}\theta}{\sec^{% 2}}\ d\theta$
33.

$\displaystyle\int_{1}^{8}\frac{2+t}{\sqrt[3]{t^{2}}}\ dt$
34.

$\displaystyle\int_{0}^{1}\sqrt[4]{x^{5}}+\sqrt[5]{x^{4}}\ dx$
35.
Explain why: (a) $\displaystyle\int_{-1}^{1}x^{n}\ dx=0$ , when $n$ is a positive, odd integer, and (b) $\displaystyle\int_{-1}^{1}x^{n}\ dx=2\int_{0}^{1}x^{n}\ dx$ when $n$ is a positive, even integer.
36.

Explain why $\displaystyle\int_{a}^{a+2\pi}\sin t\ dt=0$ for all values of $a$ .

In Exercises 37–40, find a value $c$ guaranteed by the Mean Value Theorem.

37.

$\displaystyle\int_{0}^{2}x^{2}\ dx$
38.

$\displaystyle\int_{-2}^{2}x^{2}\ dx$
39.

$\displaystyle\int_{0}^{1}e^{x}\ dx$
40.

$\displaystyle\int_{0}^{16}\sqrt{x}\ dx$

In Exercises 41–46, find the average value of the function on the given interval.

41.

$f(x)=\sin x$ on $[0,\pi/2]$
42.

$y=\sin x$ on $[0,\pi]$
43.

$y=x$ on $[0,4]$
44.

$y=x^{2}$ on $[0,4]$
45.

$y=x^{3}$ on $[0,4]$
46.

$g(t)=1/t$ on $[1,e]$

In Exercises 47–50, a velocity function of an object moving along a straight line is given. Find (a) the displacement of the object over the given time interval and (b) the total distance traveled by the object over the given time interval.

47.

$v(t)=-32t+20$ ft/s on $[0,5]$
48.

$v(t)=-32t+200$ ft/s on $[0,10]$
49.

$v(t)=\cos t$ ft/s on $[0,3\pi/2]$
50.

$v(t)=\sqrt[4]{t}$ ft/s on $[0,16]$

In Exercises 51–54, an acceleration function of an object moving along a straight line is given. Find the change of the object’s velocity over the given time interval.

51.

$a(t)=-32$ ft/s ${}^{2}$ on $[0,2]$
52.

$a(t)=10$ ft/s ${}^{2}$ on $[0,5]$
53.

$a(t)=t$ ft/s ${}^{2}$ on $[0,2]$
54.

$a(t)=\cos t$ ft/s ${}^{2}$ on $[0,\pi]$

In Exercises 55–62, use the Fundamental Theorem of Calculus Part 1 to find $F^{\prime}(x)$ .

55.

$\displaystyle F(x)=\int_{2}^{x^{3}+x}\frac{1}{t}\ dt$
56.

$\displaystyle F(x)=\int_{x^{3}}^{0}t^{3}\ dt$
57.

$\displaystyle F(x)=\int_{x}^{x^{2}}(t+2)\ dt$
58.

$\displaystyle F(x)=\int_{\ln x}^{e^{x}}\sin t\ dt$
59.

$\displaystyle F(x)=\int_{1}^{x}\frac{\ln t+4}{t^{2}+7}\ dt$
60.

$\displaystyle F(x)=\int_{2}^{\sin x}\cos^{3}t+3\tan^{3}t\ dt$
61.

$\displaystyle F(x)=\int_{5x^{3}}^{4}\frac{\sqrt{\cos t+5}}{t^{2}+e^{t}}\ dt$
62.

$\displaystyle F(x)=\int_{\tan^{2}x}^{10}\ln t+e^{t^{2}-7}\ dt$
63.
Let $g(x)=\displaystyle\int_{0}^{x}f(t)\,dt$ where $f$ is the function whose graph is shown below. (a) Evaluate $g(x)$ for $x=0,1,2,3,4,5,6$ . (b) Estimate $g(7)$ . (c) Where does $g$ have a minimum value? a maximum value? (d) Sketch the graph of $g$ .
64.
For any $x>0$ , define $g(x)=\int_{1}^{x}\frac{1}{t}\ dt.$ (a) Show that $g(x)$ is continuous and differentiable on $(0,\infty)$ with $g^{\prime}(x)=\frac{1}{x}$ . (b) Show that for any positive $x$ and $y$ we have $g(xy)=g(x)+g(y)$ . [Hint: Treat $y$ as a constant, consider the derivative with respect to $x$ of each side of the proposed equation, and apply Theorem 5.1.1.] (c) Show that for any positive $x$ and any $r$ we have $g(x^{r})=rg(x)$ . [Hint: Consider the derivative with respect to $x$ of each side of the proposed equation and apply Theorem 5.1.1.]

	$\displaystyle F(x+h)-F(x)$	$\displaystyle=\int_{a}^{x+h}f(t)\,dt-\int_{a}^{x}f(t)\,dt\quad\text{(by the % definition of $F$)}$
		$\displaystyle=\int_{x}^{x+h}f(t)\,dt,$

	$\displaystyle m(x+h-x)\leq\int_{x}^{x+h}f(t)$	$\displaystyle\,dt\leq M(x+h-x),\quad\text{or}$
	$\displaystyle mh\leq\int_{x}^{x+h}f(t)$	$\displaystyle\,dt\leq Mh.$

	$\displaystyle m\leq-\frac{1}{h}\int_{x+h}^{x}f(t)$	$\displaystyle\,dt\leq M$
	$\displaystyle f(u)=m\leq\frac{1}{h}\int_{x}^{x+h}f(t)$	$\displaystyle\,dt\leq M=f(v)\quad\text{(using \autoref{thm:defintprop}(2))}$

	$\displaystyle\int_{0}^{1}v(t)\ dt$	$\displaystyle=\int_{0}^{1}(-32t+20)\ dt$
		$\displaystyle=-16t^{2}+20t\Big{\|}_{0}^{1}$
		$\displaystyle=4\text{ ft}.$

	$\displaystyle v(t)=-32t+20$	$\displaystyle=0$
	$\displaystyle-32t$	$\displaystyle=-20$
	$\displaystyle t$	$\displaystyle=\frac{5}{8}$

	$\displaystyle\int_{0}^{1}\left\lvert v(t)\right\rvert\ dt$	$\displaystyle=\int_{0}^{5/8}v(t)\ dt+\int_{5/8}^{1}-v(t)\ dt$
		$\displaystyle=\int_{0}^{5/8}-32t+20\ dt+\int_{5/8}^{1}32t-20\ dt$
		$\displaystyle=\frac{34}{4}=8.5\text{ ft}.$

	$\displaystyle f(c)$	$\displaystyle=\frac{\int_{a}^{b}f(x)\,dx}{b-a},\text{ or}$
	$\displaystyle f(c)(b-a)$	$\displaystyle=\int_{a}^{b}f(x)\,dx.\qed$