Average value and average rates

A black-and-white close-up of water flowing from a stainless steel faucet. — Photo by ClickerHappy on Pexels.

A faucet’s flow can vary from moment to moment. If we replace that changing flow with a steady rate, what should stay the same? Over the same time, the steady flow should deliver the same amount of water. A rate that lasts longer contributes more to that amount, so averaging the settings alone may not give the replacement we need.

This question extends beyond flow rates. For a changing temperature or density, we can also seek a constant value that preserves the total accumulated over an interval. Integration accounts for both the values and the lengths of the intervals over which they contribute, giving us a way to find that constant replacement.

When the function represents an instantaneous rate, its average connects to the familiar calculation of total change divided by elapsed time. In motion, the total we preserve matters: displacement and distance traveled lead to different averages, helping explain the distinction between average velocity and average speed.

By the end of this lesson, you should be able to:

Find and interpret an average value as a constant replacement that preserves an integral over an interval.
Connect average rates of change to averages of instantaneous rates of change.
Distinguish average velocity from average speed using displacement and distance traveled.

You’re testing a faucet by collecting water in a 100-liter container. The container starts empty, the drain is closed, and no water is lost. In a first test, the faucet delivers $$ 2 $$ liters per minute for $$ 1 $$ minute, then $$ 6 $$ liters per minute for $$ 3 $$ minutes. (We’ll ignore the brief transition between the two settings.)

After four minutes, the container holds liters.

You want to repeat the test with a steady flow for the same four minutes. What flow rate would leave the same amount of water in the container?

Someone suggests averaging the two settings: $\displaystyle \frac{2+6}2=4$ liters per minute. At that steady rate, the faucet would add liters in four minutes. Compared with the original test, this is $$ 4 $$ liters too little.

The two settings last for different amounts of time. To give each minute the same weight, we can average the rates for the four one-minute intervals:

$Choose one ``\frac{2\cdot1+6\cdot3}{4}`` ``\frac{2\cdot3+6\cdot1}{4}``$\displaystyle = $$\displaystyle \text{ L/min}.$$

The longer stage contributes more water because its rate lasts longer. Each product in the numerator gives the volume added during one stage; dividing their total by four minutes gives the steady rate that would add the same volume in the same time.

We can see why this works in the graph. The original flow gives two rectangles whose combined area represents the $$ 20 $$ liters added. A steady flow gives one rectangle with the same four-minute base. To preserve the volume added, that rectangle must also have an area of $$ 20 $$ liters.

Dividing the total volume by the elapsed time finds the rectangle’s area divided by its base. Thus $$ 5 $$ is the of that rectangle.

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The two original regions and the replacement rectangle share a four-minute base and an area representing

\( 20 \)

liters added. The rectangle’s height is

\( 5 \)

liters per minute.

For a second test, the container starts empty again. This time the faucet’s flow rate is modeled by

$q(t)=t^2+1\ \text{L/min},\qquad0\le t\le3,$

where $$ t $$ is the number of minutes since this test began. The rate increases throughout the test, so a list of two settings no longer describes the flow. How could we find a steady replacement now?

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The graph still gives us a way to describe the replacement: a rectangle with the same three-minute base and the same area as the region under the curve. Its height will be the steady flow rate, but first we need the total volume added.

Over a short time interval beginning at $$ t $$ , we can treat the flow as nearly constant. During that interval of length $\Delta t$ , the volume increase $\Delta V$ is approximately liters. If $$ V $$ is the volume in the container, the relation for an infinitesimal time interval is $$\dd{V}=$Choose one `q(t)\dd{t}` ``\frac{q(t)}{\dd{t}}`` ``\frac{\dd{t}}{q(t)}``$ . Accumulating these infinitesimal contributions across the test gives the volume after three minutes, since the container started empty:

$V(3)=\int_0^3(t^2+1)\dd{t}.$

To evaluate this integral, we use an antiderivative and the endpoint-evaluation method from Section 3.3:

$\displaystyle \int_0^3 (t^2+1)\dd{t}\)$\displaystyle = $Choose one ``t^3+t`` ``t^3`` ``\tfrac13 t^3+t`` ``\tfrac13 t^3``$\displaystyle \Big|_0^3$\(\displaystyle =$

Over the three minutes, the faucet therefore adds $$ 12 $$ liters.

With the total volume known, we can find the average flow rate. We’ll denote it by $\overline{q}$ , read “q bar.” To add the same volume in three minutes, the steady rate must be $\overline{q}$ = liters per minute.

The graph shows this replacement as a rectangle three minutes wide, with height equal to the average flow rate. Its area represents the volume added, so it matches the area under the original flow-rate curve.

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A steady flow of

\( 4 \)

liters per minute and the changing flow

\( q(t)=t^2+1 \)

each add

\( 12 \)

liters in three minutes.

In both tests, we found the average flow rate by dividing the total volume added by the test duration. The first total came from two stages of constant flow; the second came from integrating a continuously changing rate. In either case, the average is the steady rate that would add the same volume in the same time. Geometrically, it’s the height of a rectangle with the same base and area as the region under the flow-rate curve.

Average value

The word “average” is commonly traced to Arabic ʿawāriyya, meaning “damaged goods,” through Italian avaria and French avarie. In English, it described sharing losses from damage to a ship or its cargo. Those losses were shared among owners or insurers in proportion to their interests. From this practice of distributing a loss, the meaning broadened to what we now know as the arithmetic mean: a total spread evenly across a group.

When we average a list of numbers, we ask: what number could replace every number in the list while keeping their total the same? For $$ 1,1, $$ and $$ 4 $$ , the total is $$ 6 $$ , so sharing that total evenly among three numbers gives $\displaystyle \frac{6}{3}=2$ :

$1+1+4\ =\ 6\ =\ 2+2+2\text.$

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The same six unit blocks form piles of

\( 1,1, \)

and

\( 4 \)

before equal sharing, and piles of

\( 2,2, \)

and

\( 2 \)

afterward.

We can ask the same question about a function $$ f $$ on $a\le x\le b$ (where $$ a<b $$ ): what constant function could replace it while keeping the integral the same? If that constant is $$ m $$ , we need

$\int_a^b m\dd{x}=\int_a^b f(x)\dd{x}.$

The constant contributes the same amount per unit of input across the whole interval. Its integral is therefore $$ m(b-a) $$ , the constant value multiplied by the interval’s length:

$m(b-a)=\int_a^b f(x)\dd{x}.$

To find the constant itself, we divide the accumulated total by the interval’s length. This plays the same role as dividing by the number of entries when we average a list: it tells us how much of the total to assign to each unit of input.

For a nonnegative function, the region under its graph and the rectangle of height $\overline{f}$ have the same base and area. The equality also works for functions with negative values: both integrals preserve the same signed accumulation.

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The varying function and its constant replacement have equal integrals: the shaded regions have the same area and the same base from

\( a \)

\( b \)

Because the average replaces the function’s values, it has the same units as those values. Integration introduces the units of the input, and dividing by the interval’s length removes them again. For a flow rate in liters per minute, the integral gives a volume in liters; dividing by the elapsed minutes gives a rate in liters per minute.

Example

A rod is $$ 2 $$ centimeters long. At distance $$ x $$ centimeters from one end, its density is ${\rho(x)=9-x^3}$ grams per centimeter. What uniform density would give a rod of the same length the same mass?

Solution

The density tells us how much mass there is per centimeter, and it varies along the rod. Integrating it over the full length gives the mass that our uniform replacement must preserve:

$\begin{aligned}\int_0^2(9-x^3)\dd{x}&=\eval{9x-\frac{x^4}{4}}_0^2\\ &=18-4\\&=14\,\text{g}.\end{aligned}$

To find the average density, we now spread those $\displaystyle 14$ grams evenly over the rod’s $\displaystyle 2$ centimeters:

$\begin{aligned}\overline{\rho}&=\frac12\int_0^2(9-x^3)\dd{x}\\ &=\frac{14}{2}\\ &=7\,\text{g/cm}.\end{aligned}$

A uniform rod with density $\displaystyle 7$ grams per centimeter has the same mass:

$\int_0^2 7\dd{x}=14\,\text{g}=\int_0^2(9-x^3)\dd{x}.$

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The varying density and the constant density of

\( 7 \)

grams per centimeter each give a mass of

\( 14 \)

grams over the same length.

The replacement does not reproduce how the mass is distributed along the rod; it preserves the total mass over the same length. The original density decreases from $\displaystyle 9$ to $\displaystyle 1$ gram per centimeter, and our average lies between them.

Average rate of change

We’ve used the quotient $\displaystyle \frac{f(b)-f(a)}{b-a}$ to describe an average rate of change since Chapter 1. It uses just two values of $$ f $$ . How can it capture an average of all the instantaneous rates between them?

The function we need to average is $$ f' $$ , whose values are the instantaneous rates of change of $$ f $$ . Its integral accumulates those rates into the change in the original quantity. We write its average as $\overline{f'}$ , placing the bar over the whole derivative. The constant-replacement idea explains why this average can be found from the endpoints.

Theorem

Average rate of change revisited

If $$ f $$ has a derivative on $a\le x\le b$ , where $$ a<b $$ , then the average value of $$ f' $$ is the average rate of change of $$ f $$ over that interval:

$\overline{f'}=\frac{f(b)-f(a)}{b-a}.$

Proof

The constant average rate $\overline{f'}$ preserves the integral of the actual rate $$ f' $$ :

$\int_a^b\overline{f'}\dd{x}=\int_a^b f'(x)\dd{x}.$

On the left, a constant rate accumulates to that rate multiplied by the interval’s length. On the right, the Fundamental Theorem of Calculus gives the change in $$ f $$ , since $$ f $$ is an antiderivative of $$ f' $$ . Hence we have

$\overline{f'}(b-a)=f(b)-f(a).$

Dividing both sides by $$ b-a $$ , we get

$\overline{f'}=\frac{f(b)-f(a)}{b-a}\text.$

Thus replacing all the instantaneous rates by the average rate produces the same total change over the same interval.

Example

A room’s temperature is modeled by $$ T(t)=18+3t^2 $$ degrees Celsius during the first two hours of a heating test. Find its average temperature and its average rate of warming.

Solution

For the average temperature, we want one temperature to represent the whole two-hour test. We therefore integrate $\displaystyle T$ itself and divide by the duration:

$\begin{aligned}\overline{T}&=\frac12\int_0^2(18+3t^2)\dd{t}\\ &=\frac12\eval{18t+t^3}_0^2\\ &=22\,{}^\circ\mathrm{C}.\end{aligned}$

The average warming rate answers a different question: what constant rate of warming would produce the same temperature increase in two hours? The theorem lets us use the endpoint temperatures:

$\begin{aligned}\overline{T'}&=\frac{T(2)-T(0)}2\\ &=\frac{30-18}2\\ &=6\,{}^\circ\mathrm{C}/\text{h}.\end{aligned}$

We can also average the instantaneous warming rate $\displaystyle T'(t)=6t$ directly: $\displaystyle \frac12\int_0^2 6t\dd{t}=6\,{}^\circ\mathrm{C}/\text{h}$ . These two calculations of the average warming rate preserve the same temperature increase over the test, as the theorem guarantees. The average temperature describes how warm the room was over that time; the average warming rate describes how quickly its temperature increased overall.

A changing rate need not stay positive. When it becomes negative, the quantity decreases, so those contributions subtract from the accumulated change. The constant average rate preserves this net change, even if it does not reproduce the increases and decreases along the way.

Example

Water enters a container at $$ 4 $$ liters per minute and drains through a separate outlet at $$ t^2 $$ liters per minute for $0\le t\le3$ , where $$ t $$ is time in minutes. There are no other sources or losses. Find the average net rate of change of the volume over these three minutes.

Solution

Both physical flow rates are nonnegative. Subtracting outflow from inflow gives the signed net rate $\displaystyle r(t)=4-t^2$ , so $\displaystyle V'(t)=r(t)$ . Accumulating this rate gives the net change in volume:

$\begin{aligned}V(3)-V(0)&=\int_0^3(4-t^2)\dd{t}\\ &=\eval{4t-\frac{t^3}{3}}_0^3\\ &=12-9\\ &=3\,\text{L}.\end{aligned}$

Dividing that net increase by the three-minute duration gives the steady net rate that would leave the same final volume:

$\begin{aligned}\overline{r}&=\frac{V(3)-V(0)}{3-0}\\ &=\frac33\\ &=1\,\text{L/min}.\end{aligned}$

The rate is zero at $\displaystyle t=2$ . Inflow exceeds outflow during the first two minutes; outflow exceeds inflow during the last minute, when $\displaystyle 4-t^2<0$ . Separating those intervals makes the contributions visible:

$\begin{aligned}\int_0^2(4-t^2)\dd{t}&=\frac{16}{3}\,\text{L},\\ \int_2^3(4-t^2)\dd{t}&=-\frac73\,\text{L}.\end{aligned}$

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The positive and negative contributions leave a net increase of

\( 3 \)

liters. A steady inflow of

\( 1 \)

liter per minute produces the same net change over three minutes.

The negative integral over the last minute means the volume decreases by $\displaystyle \frac{7}{3}$ liters because more water leaves than enters. The earlier net gain is larger, so the volume stays at or above its starting value throughout the test. A constant net rate of $\displaystyle 1$ liter per minute would produce the same increase over three minutes.

Inequalities

Sometimes we can say which accumulated total is larger without evaluating either integral. This is the reasoning behind underestimates and overestimates: if we replace every contribution by a smaller one, the resulting total cannot be larger.

Suppose $f(x)\le g(x)$ on $a\le x\le b$ , with $$ a<b $$ . Over the same small positive width $\dd{x}$ , the contributions satisfy $f(x)\dd{x}\le g(x)\dd{x}$ . Each pair of contributions uses the same width, so their sizes reflect the comparison of the function values. Accumulating them across the interval preserves that comparison:

In the picture, $\displaystyle f(x)=1+\frac{x}{2}$ and $\displaystyle g(x)=2+\frac{x^2}{2}$ on $0\le x\le2$ . The graph of $$ g $$ stays above the graph of $$ f $$ , so each vertical strip under $$ f $$ fits inside the strip under $$ g $$ .

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Both integrals use the same interval. The blue region represents the integral of

\( f \)

; adding the green region gives the integral of

\( g \)

The picture uses positive values, but the comparison also works for signed contributions. A more negative contribution is smaller than a less negative one, just as a smaller positive contribution is smaller than a larger one. Comparing the contributions in this way compares their accumulated totals.

Average velocity and average speed

We often use “velocity” and “speed” interchangeably in everyday language. To understand their different meanings in calculus, we first need to distinguish displacement from distance traveled.

For motion along a line, velocity records direction as well as how fast an object moves. If its position is $$ x(t) $$ , its velocity is $$ v(t)=x'(t) $$ , and its speed is $$ |v(t)| $$ . A positive velocity contributes motion in the positive direction; a negative velocity contributes motion in the opposite direction. Integrating velocity therefore gives displacement. Taking the absolute value counts both directions positively, so integrating speed gives distance traveled.

Over $a\le t\le b$ , where $$ a<b $$ , we therefore have

$\begin{aligned}\text{average velocity}=\frac{\text{displacement}}{b-a}&=\frac1{b-a}\int_a^b v(t)\dd{t},\\ \text{average speed}=\frac{\text{distance}}{b-a}&=\frac1{b-a}\int_a^b|v(t)|\dd{t}.\end{aligned}$

Both averages divide by the same elapsed time, but their numerators preserve different totals. If an object turns around, the return trip cancels some of its earlier displacement while adding to its distance traveled.

Example

The graph shows a cyclist’s velocity for $0\le t\le6$ , with time in minutes and velocity in miles per minute. Find the cyclist’s displacement, distance traveled, average velocity, and average speed over these six minutes.

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The cyclist reverses direction at four minutes.

Solution

Before $$ t=4 $$ , the graph lies above the axis. Its regions are a triangle with base $$ 2 $$ and height $$ 0.2 $$ , a rectangle with width $$ 1 $$ and height $$ 0.2 $$ , and a triangle with base $$ 1 $$ and height $$ 0.2 $$ . Their areas are $$ 0.2 $$ , $$ 0.2 $$ , and $$ 0.1 $$ miles.

After $$ t=4 $$ , there are two triangles below the axis, each with base $$ 1 $$ and height $$ 0.2 $$ , so each contributes $$ -0.1 $$ miles to displacement:

$\begin{aligned}\text{displacement}&=0.2+0.2+0.1-0.1-0.1\\ &=0.3\,\text{mi}.\end{aligned}$

To find distance, reflect the negative portion above the axis to obtain the speed graph. All five areas count positively:

$\begin{aligned}\text{distance traveled}&=0.2+0.2+0.1+0.1+0.1\\ &=0.7\,\text{mi}.\end{aligned}$

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The blue velocity graph gives signed areas; reflecting its negative portion gives the dashed red speed graph, whose areas all count positively.

We now have the two totals that the averages must preserve. Dividing each by the same six-minute duration gives:

$\begin{aligned}\text{average velocity}&=\frac{0.3}{6}\\ &=\frac1{20}\,\text{mi/min},\\ \text{average speed}&=\frac{0.7}{6}\\ &=\frac7{60}\,\text{mi/min}.\end{aligned}$

The cyclist ends $$ 0.3 $$ miles in the positive direction from the starting point, but travels $$ 0.7 $$ miles altogether.

The cyclist’s return trip reduced the displacement without reducing the distance traveled, so the average speed is greater than the average velocity. This comparison follows from the rates themselves, and integral comparison lets us extend it beyond this example.

Since $|v(t)|\ge v(t)$ , comparing the integrals over the same time interval gives

$\int_a^b|v(t)|\dd{t}\ge\int_a^b v(t)\dd{t}.$

Dividing by the positive duration $$ b-a $$ gives

$\text{average speed}\ge\text{average velocity}.$

Distance counts motion in either direction positively, while contributions in opposite directions can cancel in displacement. If velocity is nonnegative throughout the interval, speed equals velocity, so their averages are equal.

A pump delivers $${Q}\,\text{L}$ at $${u}\,\text{L/min}$ , followed by $${Q}\,\text{L}$ at $${v}\,\text{L/min}$ . There is no pause between stages.
1. The stages last $\text{min}$ and $\text{min}$ . Find the average flow rate for the whole delivery: $\text{L/min}$ .
  
  Solution
  The stages take $\displaystyle \frac{${Q}}{${u}}=${first_time}\,\text{min}$ and $\displaystyle \frac{${Q}}{${v}}=${second_time}\,\text{min}$ . The average flow rate is $\displaystyle \frac{${volume}}{${first_time}+${second_time}}=${average}\,\text{L/min}$ .
2. Someone reports $${wrong}\,\text{L/min}$ by averaging the two rates. Choose the assumption behind that calculation: . Explain why that assumption matters.
  
  Solution
  The ordinary mean assumes equal times at the two rates. Here the slower rate lasts $$ ${k} $$ times as long, so it receives $$ ${k} $$ times the weight. The weighted calculation is $\displaystyle \frac{${u}(${first_time})+${v}(${second_time})}{${time}}=${average}$ .
A room’s temperature $$ T(t) $$ was recorded over $$ ${duration} $$ hours, with $$ t $$ measured in hours. Assume the temperature changes monotonically between each pair of consecutive readings; it may change from warming to cooling between intervals.

$$ t $$ (h) 0 ${time1} ${time2} ${duration}

$$ T(t) $$ (°C) ${T0} ${T1} ${T2} ${T3}
1. Using one rectangle per interval between readings, the lower integral sum is and the upper integral sum is , in ${}^\circ\mathrm{C}\cdot\text{h}$ .
  
  Solution
  
  The interval widths are $$ ${d1} $$ , $$ ${d2} $$ , and $$ ${d3} $$ hours. Monotonicity makes the smaller endpoint temperature a lower bound on each interval and the larger endpoint temperature an upper bound:
  
  $\begin{aligned}\text{lower sum}&=${low1}(${d1})+${low2}(${d2})+${low3}(${d3})\\ &=${lower_sum}\,{}^\circ\mathrm{C}\cdot\text{h},\\ \text{upper sum}&=${high1}(${d1})+${high2}(${d2})+${high3}(${d3})\\ &=${upper_sum}\,{}^\circ\mathrm{C}\cdot\text{h}.\end{aligned}$
2. For $0\le t\le${duration}$ , the average temperature lies between and ${}^\circ\mathrm{C}$ . Average these bounds to give an overall estimate: ${}^\circ\mathrm{C}$ . Why is this different from averaging the four table entries?
  
  Solution
  
  Dividing the integral bounds by the positive duration gives
  
  $\frac{${lower_sum}}{${duration}}\le\overline{T}\le\frac{${upper_sum}}{${duration}}.$
  
  The midpoint of these bounds is
  
  $\begin{aligned}\text{estimated average}&=\frac12\left(\frac{${lower_sum}}{${duration}}+\frac{${upper_sum}}{${duration}}\right)\\ &=${estimate}\,{}^\circ\mathrm{C}.\end{aligned}$
  
  This estimate accounts for the different interval widths instead of giving every table entry equal weight. The midpoint is a reasonable estimate within the bounds, but it is not guaranteed to be closer to the true average than either bound. The temperatures between readings are unknown, so the estimate need not equal the exact average.
For each part, find the requested average and show the accumulation that it preserves.
1. Find the average value of $$ f(x)=${f} $$ on $0\le x\le${L}$ : .
  
  Solution
  $\begin{aligned}\overline{f}&=\frac{1}{${L}}\int_0^{${L}}${f}\dd{x}\\ &=\frac{1}{${L}}\eval{${antiderivative}}_0^{${L}}\\ &=${average}\text.\end{aligned}$ The constant $$ ${average} $$ over a width of $$ ${L} $$ preserves the integral $$ ${accumulation} $$ .
2. Find the average value of $$ g(x)=${g} $$ on $${a}\le x\le${b}$ : .
  
  Solution
  The interval length is $${b}-${parenthesize_negative(a)}=${h}$ . Thus $\begin{aligned}\overline{g}&=\frac{1}{${h}}\eval{${antiderivative}}_{${a}}^{${b}}\\ &=\frac{${upper}-${parenthesize_negative(lower)}}{${h}}\\ &=${average}\text.\end{aligned}$ Multiplying by $$ ${h} $$ recovers the integral $$ ${accumulation} $$ .
3. A $${L}\,\text{m}$ cable has density $\rho(x)=${rho}\,\text{kg/m}$ , where $$ x $$ is distance in meters from one end and $0\le x\le${L}$ . Find its mass, $\text{kg}$ , and its average density, $\text{kg/m}$ . Explain what a uniform cable with that density would preserve.
  
  Solution
  The mass is $\displaystyle \eval{${antiderivative}}_0^{${L}}=${mass}\,\text{kg}$ . Dividing by $${L}\,\text{m}$ gives $\displaystyle \overline{\rho}=${average}\,\text{kg/m}$ . A uniform cable of the same length and this density also has mass $${mass}\,\text{kg}$ . The density ranges from $$ ${C} $$ to $${maximum}\,\text{kg/m}$ , consistent with the average.
During a $${L}\,\text{min}$ test, the volume in a $200\,\text{L}$ container is modeled by $V(t)=${V}\,\text{L}$ , for $0\le t\le${L}$ .
1. Find the average volume, $\overline{V}=$ $\text{L}$ .
  
  Solution
  $\begin{aligned}\overline{V}&=\frac1{${L}}\int_0^{${L}}(${V})\dd{t}\\ &=\frac1{${L}}\eval{${antiderivative}}_0^{${L}}\\ &=${average_volume}\,\text{L}\end{aligned}$ .
2. Find the average rate of change of volume from its endpoints: $\text{L/min}$ . Then verify this result by averaging $$ V'(t) $$ .
  
  Solution
  The endpoint quotient is $\displaystyle \frac{${final}-${C}}{${L}}=${average_rate}\,\text{L/min}$ . Since $$ V'(t)=${derivative} $$ , averaging the derivative gives $\begin{aligned}\frac1{${L}}\int_0^{${L}}${derivative}\dd{t}&=\frac1{${L}}\eval{${rate_antiderivative}}_0^{${L}}\\ &=${average_rate}\,\text{L/min}\end{aligned}$ .
3. Suppose the test instead starts with $${shifted_start}\,\text{L}$ and follows the model $$ W(t)=${W} $$ . Find the new average volume, $\text{L}$ , and average rate of change, $\text{L/min}$ . Explain why changing the starting volume affects one average but not the other.
  
  Solution
  The new average volume is $${shifted_average}\,\text{L}$ , and the average rate is still $${average_rate}\,\text{L/min}$ . The vertical shift ${shift_direction} every volume, and its average, by $${shift_size}\,\text{L}$ . The added constant cancels in the endpoint difference, and its derivative is zero. Both models keep a positive stored volume throughout the test; a negative average rate means the volume decreases, not that the average volume is negative.
A runner moves along a straight path for $0\le t\le${duration}$ seconds. The graph shows signed velocity and three regions with geometric areas $$ A=${A} $$ , $$ B=${B} $$ , and $$ C=${C} $$ . Use these supplied areas; you do not need to calculate them from the graph.

Loading the graph…If it does not appear, check the Desmos API configuration.

The supplied areas are positive geometric areas; a region below the axis contributes negatively to displacement.
1. Over the full $$ ${duration} $$ seconds, the displacement is $\text{m}$ and the distance traveled is $\text{m}$ . Explain how a region below the axis contributes to each.
  
  Solution
  
  For displacement, use a negative contribution for every region below the axis. For distance, count all three geometric areas positively:
  
  $\begin{aligned}\text{displacement}&=${signed_A}+${parenthesize_negative(signed_B)}+${parenthesize_negative(signed_C)}\\ &=${displacement}\,\text{m},\\ \text{distance traveled}&=${A}+${B}+${C}\\ &=${distance}\,\text{m}.\end{aligned}$
  
  Loading the graph…If it does not appear, check the Desmos API configuration.
  
  The dashed red speed graph counts motion in either direction positively.
2. The average velocity is $\text{m/s}$ and the average speed is $\text{m/s}$ .
  
  Solution
  
  Divide each total by the same $$ ${duration} $$ -second duration:
  
  $\begin{aligned}\text{average velocity}&=\frac{${displacement}}{${duration}}\\ &=${average_velocity}\,\text{m/s},\\ \text{average speed}&=\frac{${distance}}{${duration}}\\ &=${average_speed}\,\text{m/s}.\end{aligned}$
3. For just the first $$ ${t6} $$ seconds, covering regions A and B, the average velocity is $\text{m/s}$ and the average speed is $\text{m/s}$ . Does the runner return to the starting point at that time? Explain.
  
  Solution
  
  The signed contributions of A and B give displacement $${early_displacement}$ meters, while their geometric areas give distance $${early_distance}$ meters. ${position_note}
  
  $\begin{aligned}\text{average velocity}&=\frac{${signed_A}+${parenthesize_negative(signed_B)}}{${t6}}\\ &=${early_velocity}\,\text{m/s},\\ \text{average speed}&=\frac{${A}+${B}}{${t6}}\\ &=${early_speed}\,\text{m/s}.\end{aligned}$
Let $$ f(x)=mx+c $$ on $a\le x\le b$ , where $$ a<b $$ .
1. Find $\overline{f}$ in terms of $$ m,c,a, $$ and $$ b $$ : . Compare it with $\displaystyle f\left(\frac{a+b}{2}\right)$ and $\displaystyle \frac{f(a)+f(b)}{2}$ .
  
  Solution
  $\begin{aligned}\overline{f}&=\frac1{b-a}\eval{\frac{mx^2}{2}+cx}_a^b\\ &=\frac{m(a+b)}2+c\end{aligned}$ , using $$ b^2-a^2=(b-a)(a+b) $$ . Substituting the midpoint into $$ f $$ gives the same expression, as does taking the mean of $$ f(a) $$ and $$ f(b) $$ .
2. Now let $$ g(x)=Ax^2+Bx+C $$ for $a\le x\le b$ , where $$ a<b $$ . Use part (a) on $$ g' $$ and the average-rate theorem to show that the tangent slope at the midpoint $\displaystyle x=\frac{a+b}{2}$ equals the secant slope between the endpoints:
  
  $g'\left(\frac{a+b}{2}\right)=\frac{g(b)-g(a)}{b-a}.$
  
  Solution
  
  The derivative is $\displaystyle g'(x)=2Ax+B$ , a linear function. Part (a) shows that the average of a linear function equals its value at the midpoint. Applied to $$ g' $$ , this gives
  
  $\overline{g'}=g'\left(\frac{a+b}{2}\right).$
  
  The average-rate theorem also gives
  
  $\overline{g'}=\frac{g(b)-g(a)}{b-a}.$
  
  Both slopes equal the average of the instantaneous slopes, so
  
  $g'\left(\frac{a+b}{2}\right)=\frac{g(b)-g(a)}{b-a}.$
  
  For a quadratic position function, this says that the instantaneous velocity halfway through a time interval equals the average velocity over the interval.