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Calculus Concepts and Modelling: An introduction to the concepts of calculus

Jeremy Sylvestre

Chapter 13 Average value and the Fundamental Theorem

Section 13.1 Average value of a function

Given a quantity function \(q(t)\text{,}\) we have defined average rate of variation over a time domain \(t_1 \le t \le t_2\) as
\begin{equation*} \avg{r} = \frac{\text{accumulation}}{\text{duration}} = \slope{q}{t} \text{,} \end{equation*}
where
\begin{align*} \change{q} \amp = q(t_2) - q(t_1) \amp \change{t} \amp = t_2 - t_1 \text{.} \end{align*}
However, if \(r(t)\) is the rate function that describes the variation in \(q(t)\text{,}\) then a definite integral can also be used to measure accumulation over \(t_1 \le t \le t_2\text{:}\)
\begin{equation*} \text{accumulation} = \ccmint{t_1}{t_2}{r(t)}{t} \text{.} \end{equation*}
Substituting this into the average rate formula, we have
\begin{equation*} \avg{r} = \frac{\text{accumulation}}{\text{duration}} = \frac{1}{\change{t}} \ccmint{t_1}{t_2}{r(t)}{t} \text{.} \end{equation*}
This formula is interesting in that it relates average rate \(\avg{r}\) directly to the rate function \(r(t)\text{.}\)
What if we applied this relationship to a quantity function?

Example 13.1.1. Average temperature.

Suppose you have modelled the outdoor temperature in some location over a twenty-four hour period as
\begin{align*} T(t) \amp = \frac{t^4 - 48 t^3 + 576 t^2 + 14000}{1400} \amp 0 \amp \le t \le 24 \text{,} \end{align*}
where \(T(t)\) is expressed in degrees Celsius, \(t\) is time in hours.
A temperature graph over twenty-four hours.
Figure 13.1.2. A temperature graph over twenty-four hours.
What was the average temperature over that period? We could answer this by imagining that \(T(t)\) is not a quantity function but a rate function. Instead of being expressed in degrees, we could think of \(T(t)\) as being expressed in degree-hours per hour. (With unit analysis, we see that this made-up unit is really just degrees, as hours per hour “cancels” out.) Then we could calculate how many degree-hours accumulated over this period, using Pattern 7.4.4 to assist with the calculation.
\begin{align*} \ccmint{0}{24}{T(t)}{t} \amp = \ccmint{0}{24}{\frac{t^4 - 48 t^3 + 576 t^2 + 14000}{1400}}{t}\\ \amp = \ccmint{0}{24}{\frac{t^4}{1400} - \frac{6 t^3}{175} + \frac{72 t^2}{175} + 10}{t} \\ \amp = \frac{{24}^5}{5 \cdot 1400} - \frac{6 \cdot {24}^4}{4 \cdot 175} + \frac{72 {24}^3}{3 \cdot 175} + 10 \cdot 24 \\ \amp = \frac{375888}{875} \end{align*}
To get the average “rate”, we divide by the duration:
\begin{align*} \avg{T} \amp = \frac{\text{accumulation}}{\text{duration}} \\ \amp = \frac{375888/875}{24} \\ \amp = 15662/875 \\ \amp \approx 17.9 \text{.} \end{align*}
But this isn’t really a rate, we were only pretending it was a rate in the made-up units of degree-hours per hour. Ignoring the hours-per-hour, it seems that the average temperature over that twenty-four hour period was approximately 17.9 °C.
Average temperature over a twenty-four-hour period.
Figure 13.1.3. Average temperature over a twenty-four-hour period.

Remark 13.1.4.

In Example 13.1.1, it was correct to say that when we calculated
\begin{equation*} \ccmint{0}{24}{T(t)}{t} \text{,} \end{equation*}
we were calculating a quantity in units of degree-hours. Consider a (regular) Riemann sum to approximate that definite integral:
\begin{align*} \ccmint{0}{24}{T(t)}{t} \amp \approx \sum_{k=1}^n T(t_k^\ast) \change{t} \\ \amp = T(t_1^\ast) \change{t} + T(t_2^\ast) \change{t} + \dotsb + T(t_n^\ast) \change{t} \text{.} \end{align*}
In each term of the sum, we are multiplying a temperature \(T(t_k^\ast)\) in degrees against a small time duration \(\change{t}\) in hours, so the product \(T(t_k^\ast) \change{t}\) is in units of degree-hours.
And then when we calculated \(\avg{T}\text{,}\) we divided our accumulation in degree-hours by the full duration of 24 h, so that this quotient ended up in units of degrees Celsius.

Definition 13.1.5. Average value of a function.

Given integrable function \(f(t)\) over domain \(a \le t \le b\text{,}\) we define the average value of \(f\) on that domain to be
\begin{equation*} \avg{f} = \frac{1}{b - a} \ccmint{a}{b}{f(t)}{t} \text{.} \end{equation*}
Graphically, if we think of the definite integral in this definition as an area, then we are dividing that area by the full width \(b - a\) to obtain an average height:
\begin{equation*} \text{area} = \text{height} \times \text{width} \qquad\implies\qquad \text{height} = \frac{\text{area}}{\text{width}} \text{.} \end{equation*}
But we should also compare our definition of Average value of a function with our definition of Average rate of accumulation — what if we apply our definition above to the case of a rate function \(r(t)\text{?}\) Will we be calculating average rate of accumulation as we’ve defined it in Definition 11.1.2? Our definition of average rate is that it is the constant rate that produces the same accumulation as the varying rate function over a given time period. For a constant rate \(r = \avg{r}\) we have
\begin{equation*} \text{accumulation} = \text{rate} \times \text{duration} = \avg{r} \times (b - a) \text{.} \end{equation*}
For a varying rate \(r(t)\) we have
\begin{equation*} \text{accumulation} = \ccmint{a}{b}{r(t)}{t} \text{.} \end{equation*}
But if \(\avg{r}\) is taken to be
\begin{equation*} \avg{r} = \frac{1}{b - a} \ccmint{a}{b}{r(t)}{t} \end{equation*}
as in Definition 13.1.5, then the two accumulation formulas above become the same.
Graph demonstrating that Average temperature over a twenty-four-hour period achieves the same total accumulation of degree-hours as measured by the integral.
Figure 13.1.6. Average temperature over a twenty-four-hour period achieves the same total “accumulation” of degree-hours as measured by the integral.
Compare Figure 13.1.6 with the graphical interpretation we made of average rate in Section 11.2.

Example 13.1.7. Calculating average value of a function.

What is the average value of the cosine function over the domain \(-\pi/2 \le t \le \pi/2\text{?}\) First let’s calculate the accumulation
\begin{align*} \ccmint{-\pi/2}{\pi/2}{\cos(t)}{t} \amp = \ccmint{-\pi/2}{0}{\cos(t)}{t} + \ccmint{0}{\pi/2}{\cos(t)}{t}\\ \amp = - \ccmint{0}{-\pi/2}{\cos(t)}{t} + \ccmint{0}{\pi/2}{\cos(t)}{t} \\ \amp = - \sin(-\pi/2) + \sin(\pi/2) \\ \amp = - (-1) + 1 \\ \amp = 2 \text{.} \end{align*}
So, using Definition 13.1.5, over \(-\pi/2 \le t \le \pi/2\) we have
\begin{align*} \avg{\cos} \amp = \frac{1}{\frac{\pi}{2} - (- \frac{\pi}{2})} \ccmint{-\pi/2}{\pi/2}{\cos(t)}{t}\\ \amp = \frac{1}{\frac{\pi}{2} + \frac{\pi}{2}} \cdot 2 \\ \amp = \frac{2}{\pi} \text{.} \end{align*}
Finally, let’s compare this integral-based notion of average value with our previous notion: given values \(x_1,x_2,\dotsc,x_n\text{,}\) the average of these values is defined to be
\begin{equation*} \frac{x_1 + x_2 + \dotsb + x_n}{n} \text{.} \end{equation*}

Example 13.1.8. Computing average temperature from temperature samples.

Let’s return to the situation of Example 13.1.1, where \(T(t)\) represents the temperature in some outdoor location at time \(t\) over a twenty-four hour period, but this time suppose we haven’t modelled \(T(t)\) with a formula. We could approximate \(\avg{T}\) by making temperature measurements throughout the day, perhaps at the end of every hour:
\begin{align*} T_1 \amp = T(1) \amp T_2 \amp = T(2) \amp \amp \cdots \amp T_{24} \amp = T(24)\text{.} \end{align*}
Then we would calculate
\begin{equation*} \avg{T} \approx \frac{T_1 + T_2 + \dotsb + T_{24}}{24} \text{.} \end{equation*}
To get a better approximation of the average, we might connect a digital thermometer to a computer and record temperatures every minute, so that
\begin{equation*} \avg{T} \approx \frac{T_1 + T_2 + \dotsb + T_{1440}}{1440} \text{,} \end{equation*}
where now \(T_k = T(k/60)\) because the \(t\) in \(T(t)\) is measured in hours. More generally, if we were to record the temperature at regular intervals, \(n\) times during the day, we would calculate
\begin{gather} \avg{T} \approx \frac{T_1 + T_2 + \dotsb + T_n}{n}\text{.}\tag{✶} \end{gather}
Adding things up over shorter time intervals to get a better approximation is starting to sound like a Riemann sum. If we were to create a (regular) Riemann sum for function \(T(t)\) over domain \(0 \le t \le 24\text{,}\) we would start with
\begin{equation*} \change{t} = \frac{24 - 0}{n} = \frac{24}{n} \text{.} \end{equation*}
Let’s substitute
\begin{equation*} n = \frac{24}{\change{t}} \end{equation*}
into (✶):
\begin{align*} \avg{T} \amp \approx \frac{T_1 + T_2 + \dotsb + T_n}{24/\change{t}} \\ \amp = \frac{\change{t}}{24} (T_1 + T_2 + \dotsb + T_n) \\ \amp = \frac{1}{24} (T_1 \change{t} + T_2 \change{t} + \dotsb + T_n \change{t}) \\ \amp = \frac{1}{24} \sum_{k = 1}^n T_k \change{t} \text{.} \end{align*}
As the temperature readings \(T_k\) are essentially samples of the temperature function,
\begin{equation*} T_k = T(t_k^\ast) \text{,} \end{equation*}
where the sample points \(t_k^\ast\) might be at the end of each time interval between readings, the above summation is a right regular Riemann sum! This means we can turn the approximation into an equality by replacing the Riemann sum with an integral:
\begin{equation*} \avg{T} \approx \frac{1}{24} \sum_{k = 1}^n T(t_k^\ast) \change{t} \qquad\implies\qquad \avg{T} = \frac{1}{24} \ccmint{0}{24}{T(t)}{t}\text{.} \end{equation*}
Since \(24\) is our total time duration, this formula for \(\avg{T}\) matches exactly with Definition 13.1.5.

Remark 13.1.9.

The analysis strategy in Example 13.1.8 is a powerful method in engineering and the sciences: if you can interpret an approximation to some value as a Riemann sum, then you can calculate that value exactly with an integral.

Section 13.2 The Mean Value Theorem

The Mean Value Theorem says something inherently intuitive: given an average value \(\avg{q}\) for some quantity \(q(t)\) that is varying, it’s not possible for that quantity to have remained above average for the entire period, nor to have remained below average for the entire period. And if the function is continuous, that means it must have crossed through its average value at some point.

Justification.

Let \(\submin{t}\) be a time in \(a \le t \le b\) where \(f\) achieves its least value, and let \(\submax{t}\) be a time in \(a \le t \le b\) where \(f\) achieves its greatest value. Then over our domain, the area under the horizontal line \(y = f(\submin{t})\) must be less than the area under the graph of \(f(t)\text{.}\) In other words, if we think of \(f(t)\) as a rate function, the accumulation at the slowest rate must be less than the true accumulation. Similarly, the area under the horizontal line \(y = f(\submax{t})\) must be greater than the area under the graph of \(f(t)\text{.}\) (Or, the accumulation at the fastest rate must be greater than the true accumulation.)
Graph demonstrating that total accumulation is always greater than accumulation at the lowest rate.
(a) The total accumulation is always greater than accumulation at the lowest rate.
Graph demonstrating that total accumulation is always less than accumulation at the highest rate.
(b) The total accumulation is always less than accumulation at the highest rate.
Figure 13.2.2. Comparing accumulation at the minimum and maximum rates.
Putting these together, we have
\begin{gather*} f(\submin{t}) \cdot (b - a) \le \ccmint{a}{b}{f(t)}{t} \le f(\submax{t}) \cdot (b - a) \\ f(\submin{t}) \le \frac{1}{b - a} \ccmint{a}{b}{f(t)}{t} \le f(\submax{t}) \\ f(\submin{t}) \le \avg{f} \le f(\submax{t}) \end{gather*}
Since \(f(t)\) is assumed to be continuous, it is not possible for its values to have “jumped” over \(y = \avg{f}\) on its way from \(y = f(\submin{t})\) to \(y = f(\submax{t})\) (or vice versa, in the case that \(\submax{t}\) is earlier than \(\submin{t}\)). That is, we may apply the Intermediate Value Theorem to conclude that there is some time \(t = c\) between \(t = \submin{t}\) and \(t = \submax{t}\) where \(f(c) = \avg{f}\text{.}\)
Graph demonstrating that a continuous function must achieve its average value.
Figure 13.2.3. A continuous function must achieve its average value.

Example 13.2.4. Achieving average speed.

Suppose you leave your home and drive to a destination 100 km away, and the trip takes about an hour. Then your average speed was approximately 100 km⁄h. But you most likely were driving much slower than that at the beginning and end of the trip, as you probably had city speed limits and traffic to deal with. Which means that to have averaged 100 km⁄h, you must also have been travelling faster than that for some period, to average out with your slower speeds. From this we make a slightly stronger conclusion than that in the Mean Value Theorem: your speedometer must have read exactly 100 km⁄h at least twice during your trip, once when you accelerated from your initial slower speeds to your later higher speeds, and again when you decelerated from your later higher speeds to slower speeds near the end of your trip.

Example 13.2.5. Determining when average value was achieved.

For the function
\begin{equation*} f(t) = t (5 - t) \text{,} \end{equation*}
over the interval \(0 \le t \le 4\) we have
\begin{align*} \ccmint{0}{4}{f(t)}{t} \amp = \ccmint{0}{4}{5 t - t^2}{t} \\ \amp = 5 \cdot \frac{4^2}{2} - \frac{4^3}{3} \\ \amp = \frac{56}{3} \text{,} \end{align*}
so
\begin{equation*} \avg{f} = \frac{1}{4 - 0} \cdot \frac{56}{3} = \frac{14}{3} \text{.} \end{equation*}
The Mean Value Theorem tells us that \(f(t) = \avg{f} \) should occur at least once in the domain \(0 \le t \le 4\text{:}\)
\begin{gather*} 5 t - t^2 = \frac{14}{3} \\ 15 t - 3 t^2 = 14 \\ 3 t^2 - 15 t + 14 = 0 \text{.} \end{gather*}
Using the Quadratic formula, we obtain two solutions:
\begin{align*} t \amp = \frac{15 - \sqrt{{15}^2 - 4 \cdot 3 \cdot 14}}{2 \cdot 3} \amp t \amp = \frac{15 + \sqrt{{15}^2 - 4 \cdot 3 \cdot 14}}{2 \cdot 3}\\ \amp = \frac{15 - \sqrt{57}}{6} \amp \amp = \frac{15 + \sqrt{57}}{6}\text{.} \end{align*}
Both of these are within our domain, so the function achieves its maximum twice, once on the way up, and again on the way down.
Graph of a parabola achieving its average value twice.
Figure 13.2.6. A parabola achieving its average value twice.

Section 13.3 The Fundamental Theorem

Let’s review.
We have learned how to use a Riemann sum to approximate an accumulation from the rate function \(r(t)\text{.}\) Since the true accumulation value always lies between the upper and lower Riemann sums, for an integrable function we can “squeeze” the upper and lower sums together by using more, skinnier rectangles in our sums until the upper and lower sums are so close together that we can determine the true accumulation value. We called this value the definite integral for the rate function and the domain over which we were computing an accumulation. By using a variable endpoint for accumulation calculations, we can create an accumulation function. Using Pattern 6.4.7, we can combine an accumulation function with an initial value to recover the quantity function \(q(t)\text{.}\)
We have also learned how to use average rates of variation to approximate an instantaneous rate of variation from values of a quantity function \(q(t)\text{.}\) Over shorter and shorter time intervals for the average rate calculations, if all such approximations are arbitrarily close to some particular value, we call that value the derivative value for the quantity function at the particular instant in question, and we interpret it as the rate of variation of the quantity function at that instant. By “stitching” together derivative values at different instants together into a function called the derivative function \(q'(t)\) for the quantity function. Again, interpreting values of \(q'(t)\) as instantaneous rates of variation, it is reasonable to think that in forming the derivative function we have recovered the rate function, so that \(q'(t) = r(t)\text{.}\)
If this is the case, then performing both processes should return us to where we started: if we begin with a rate function \(r(t)\text{,}\) use an accumulation function based on a definite integral of \(r(t)\) to obtain a quantity function, and then compute the derivative of that quantity function, we should be back at the original rate function. Intuitively, this is beyond doubt. However, mathematically there is no reason to think this should be true, as the two processes involved have seemingly nothing to do with each other: one process uses rectangles to approximate an area, the other process uses secant-line slopes to approximate a tangent-line slope. Why would performing one process after the other return you to where you started? But, in fact, our intuition based on rates, quantities, and accumulation is correct.

Justification.

As usual, to compute the derivative of \(A(t)\text{,}\) we consider a forward/backward difference quotient. First consider the numerator.
\begin{equation*} A(t + dt) - A(t) = \ccmint{a}{t + dt}{f(u)}{u} - \ccmint{a}{t}{f(u)}{u} \text{.} \end{equation*}
Regardless of whether \(dt\) is positive or negative, we can use Property 5 of Pattern 6.3.11 to split that first integral up.
\begin{equation*} A(t + dt) - A(t) = \left( \ccmint{a}{t}{f(u)}{u} + \ccmint{t}{t + dt}{f(u)}{u} \right) - \ccmint{a}{t}{f(u)}{u}\text{.} \end{equation*}
But we can cancel the first and third integrals above to get
\begin{equation*} A(t + dt) - A(t) = \ccmint{t}{t + dt}{f(u)}{u} \text{.} \end{equation*}
We now have a simplified version of the numerator to use in our difference quotient:
\begin{align*} \ddt{A} \amp = \frac{A(t + dt) - A(t)}{dt} \\ \amp = \frac{1}{dt} \ccmint{t}{t + dt}{f(u)}{u} \text{.} \end{align*}
The integral is over the domain \(t \le u \le t + dt\text{,}\) and \(dt\) is the duration of that domain, which means that
\begin{equation*} \ddt{A} = \avg{f} \text{,} \end{equation*}
where the average is over that domain \(t \le u \le t + dt\text{.}\) The Mean Value Theorem says that there is at least one time \(c\) in that domain where \(f(c) = \avg{f}\text{.}\) But when \(dt \approx 0\text{,}\) then
\begin{equation*} t + dt \approx t \text{,} \end{equation*}
and so this \(c\) that is between \(t\) and \(t + dt\) must be “squeezed” down to \(c \approx t\) as well. That is,
\begin{equation*} \ddt{A} = \avg{f} = f(c) = f(t) \text{,} \end{equation*}
as desired.
Now, Pattern 11.7.9 says that a discontinuous function cannot be differentiable, so since our accumulation function \(A(t)\) is differentiable it must be continuous, at least at all points in the domain \(a \lt t \lt b\text{.}\) At the endpoints \(t = a\) and \(t = b\) it is more complicated, and we leave that justification to your further study of calculus, if you undertake it.

Remark 13.3.2.

If we accept that a definite integral does in fact measure accumulation, then Theorem 13.3.1 confirms that a derivative function is a rate function, as we had proposed in Principle 12.1.2. If \(q(t)\) is a quantity function with associated rate of variation function \(r(t)\text{,}\) then from \(r(t)\) we can recover the quantity function using Pattern 6.4.7:
\begin{equation*} q(t) = q(a) + \ccmint{a}{t}{r(u)}{u} \text{.} \end{equation*}
Then, using Theorem 13.3.1 and the properties of differentiation from Section 12.3, we can compute
\begin{align*} q'(t) \amp = \opddt \left( q(a) + \ccmint{a}{t}{r(u)}{u} \right) \\ \amp = 0 + \opddt \ccmint{a}{t}{r(u)}{u} \\ \amp = r(t) \text{.} \end{align*}

Example 13.3.3. Derivative of \(\ln(t)\) and \(\ln\abs{t}\).

We can now use Fundamental Theorem of Calculus to verify Pattern 12.5.1:
\begin{align*} \ddt{\bbrac{\ln (t)}} \amp = \opddt \ccmint{1}{t}{\frac{1}{u}}{u} \\ \amp = \frac{1}{t} \text{.} \end{align*}
We can similarly verify Pattern 12.5.4 using our interpretation of \(\ln\abs{t}\) at the end of Section 12.5:
\begin{equation*} \ln\abs{t} = \begin{cases} \displaystyle \ccmint{1}{t}{\frac{1}{u}}{u} \amp t \gt 0 \\[3pt] \displaystyle \ccmint{-1}{t}{\frac{1}{u}}{u} \amp t \lt 0 \end{cases}\text{.} \end{equation*}
Therefore,
\begin{align*} \ddt{\bbrac{\ln \abs{t}}} \amp = \begin{cases} \displaystyle \opddt \ccmint{1}{t}{\frac{1}{u}}{u} \amp t \gt 0 \\[3pt] \displaystyle \opddt \ccmint{-1}{t}{\frac{1}{u}}{u} \amp t \lt 0 \end{cases}\\ \amp = \begin{cases} \displaystyle \frac{1}{t} \amp t \gt 0 \\[3pt] \displaystyle \frac{1}{t} \amp t \lt 0 \end{cases}\\ \amp = \frac{1}{t} \text{.} \end{align*}

Example 13.3.4. Derivative of a trigonometric accumulation function.

In Section 10.5, we used the example of a spring attached to a mass as inspiration for the formulas
\begin{align*} \ccmint{0}{t}{\sin(u)}{u} \amp = 1 - \cos(t) \amp \ccmint{0}{t}{\cos(u)}{u} \amp = \sin(t)\text{.} \end{align*}
(See Pattern 10.5.1 and Pattern 10.5.2.)
Suppose we have a quantity function \(q(t)\) and associated rate function \(r(t)\text{,}\) and we have modelled \(r(t) = \sin(t)\text{.}\) As usual, we can create an accumulation function
\begin{equation*} A_0(t) = \ccmint{0}{t}{\sin(u)}{u} \text{,} \end{equation*}
where the value of \(A_0(t)\) represents the accumulation of quantity that occurred over the domain from time \(0\) up until time \(t\text{.}\) Applying the Fundamental Theorem of Calculus to compute the derivative function returns us to the rate function, as expected:
\begin{equation*} \ddt{A_0} = \opddt \ccmint{0}{t}{\sin(u)}{u} = \sin(t) \text{.} \end{equation*}
On the other hand, applying Pattern 10.5.1 tells us that
\begin{equation*} A_0(t) = 1 - \cos(t) \text{.} \end{equation*}
What happens if we differentiate this formula using our patterns from Chapter 12?
\begin{align*} \ddt{A_0} \amp = \opddt \bbrac{1 - \cos(t)} \\ \amp = \ddt{1} - \ddt{\bbrac{\cos(t)}} \\ \amp = 0 - \bbrac{- \sin(t)} \\ \amp = \sin(t) \end{align*}
It is good that we got the same result both times! While this by itself does not imply that our accumulation formula for the sine function (Pattern 10.5.1) is correct, it certainly gives us more confidence.

Section 13.4 Antiderivatives

The Fundamental Theorem of Calculus tells us that if we begin with a rate function \(r(t)\text{,}\) use a definite integral to recover the quantity function
\begin{equation*} q(t) = q_0 + \ccmint{t_0}{t}{r(u)}{u} \end{equation*}
(where \(q_0\) represents a chosen or observed “initial value” \(q(t_0) = q_0\)), and then differentiate, we end up back at the rate function:
\begin{equation*} q'(t) = r(t) \text{.} \end{equation*}
(See the calculation in Remark 13.3.2.)
What if we begin with a quantity function \(q(t)\text{?}\) Suppose we take the derivative as the rate function, \(r(t) = q'(t)\text{,}\) and then use that to re-form a quantity function. Except at first, there is not yet a mathematical reason to expect that we will end up back at \(q(t)\text{,}\) so let’s write
\begin{equation*} s(t) = s_0 + \ccmint{t_0}{t}{r(u)}{u} \end{equation*}
for this quantity function, where again \(s_0\) represents a chosen “initial value” \(s(t_0) = s_0\text{.}\) How is \(s(t)\) related to \(q(t)\text{?}\) We at least know that they have the same derivative function: using an identical calculation as in Remark 13.3.2, differentiating the formula for \(s(t)\) above arrives at
\begin{equation*} s'(t) = r(t) = q'(t) \text{.} \end{equation*}

Definition 13.4.1. Antiderivative.

If \(f(t)\) and \(F(t)\) are functions so that \(F(t)\) is differentiable with \(F'(t) = f(t)\text{,}\) then \(F(t)\) is called an antiderivative of \(f(t)\text{.}\)

Remark 13.4.2.

For \(f(t)\) and \(F(t)\) in Definition 13.4.1, while we say that \(F'(t) = f(t)\) is the derivative of \(F(t)\text{,}\) we are careful to say that \(F(t)\) is an antiderivative of \(f(t)\text{,}\) because in fact a function that has an antiderivative always has an infinite number of antiderivatives.

Example 13.4.3. Two antiderivatives for the same function.

Consider \(F_1(t) = t^2\) and \(F_2(t) = t^2 + 1\text{.}\) Using the patterns in Chapter 12, we have both
\begin{align*} F_1'(t) \amp = 2 t \amp F_2'(t) \amp = 2 t + 0 = 2 t \text{.} \end{align*}
This tells us that both \(F_1\) and \(F_2\) are antiderivatives of the function \(f(t) = 2 t\text{.}\) And because the derivative of a constant is \(0\text{,}\) we could change the plus-one part of the formula for \(F_2(t)\) to be any other constant and it would still be an antiderivative for \(f(t)\text{.}\)
What functions have antiderivatives? That is, what functions are derivative functions? That is a big question, but the Fundamental Theorem of Calculus provides an important partial answer.

Example 13.4.5. Accumulation of a line.

The function \(f(t) = t\) is obviously continuous. Using simple geometry (see Example 7.3.1) we can compute
\begin{align*} A_0(t) \amp = \ccmint{0}{t}{f(u)}{u} \\ \amp = \ccmint{0}{t}{u}{u} \\ \amp = \frac{t^2}{2} \text{.} \end{align*}
Using Pattern 12.4.3 we can compute
\begin{align*} A_0'(t) \amp = \opddt \left( \frac{t^2}{2} \right) \\ \amp = \frac{2 t}{2} \\ \amp = t \\ \amp = f(t) \text{.} \end{align*}
So treating our original function \(f(t)\) as a rate function, the associated accumulation function is an antiderivative.
Now we return to the question that began the section, where we created a function that had the same derivative as another, and asked how the two functions were related. First, we can make the pattern of Example 13.4.3 more general.

Justification.

Consider the difference function
\begin{equation*} G(t) = F_2(t) - F_1(t) \text{.} \end{equation*}
Then
\begin{align*} G'(t) \amp = F_2'(t) - F_1'(t) \\ \amp = f(t) - f(t) \\ \amp = 0 \text{.} \end{align*}
As a derivative function is a rate function, having \(G'(t)\) uniformly equal to \(0\) says that the function \(G(t)\) never varies, so it must be a constant function \(G(t) = C\text{.}\) As \(G\) was defined as the difference of \(F_1\) and \(F_2\text{,}\) we have
\begin{equation*} F_2(t) - F_1(t) = C \qquad\implies\qquad F_2(t) = F_1(t) + C \text{,} \end{equation*}
so that \(F_2(t)\) is a vertical shift of \(F_1(t)\text{.}\) (Alternatively, we could have isolated \(F_1(t)\) as a vertical shift of \(F_2(t)\text{.}\))

Justification.

Suppose \(q_1(t)\) and \(q_2(t)\) are two functions so that \(q_1'(t) = q_2'(t)\text{.}\) If we write \(f(t)\) to represent this shared derivative function, then both \(q_1\) and \(q_2\) are antiderivatives of that function, and we are in the situation of Fact 13.4.6 with \(q_1\) in place of \(F_1\) and \(q_2\) in place of \(F_2\text{.}\)
What about two different accumulation functions? If the function \(f(t)\) is continuous on a large domain, when creating an accumulation antiderivative as in Fact 13.4.4, we can choose any base point \(a\) for the lower bound of our definite integral that we like. So, starting with continuous \(f(t)\text{,}\) we can create two different antiderivatives
\begin{align*} A_1(t) \amp = \ccmint{a_1}{t}{f(u)}{u} \amp A_2(t) \amp = \ccmint{a_2}{t}{f(u)}{u}\text{.} \end{align*}
But these are still vertical shifts of one another, since
\begin{align*} A_1(t) \amp = \ccmint{a_1}{t}{f(u)}{u} \\ \amp = \ccmint{a_1}{a_2}{f(u)}{u} + \ccmint{a_2}{t}{f(u)}{u} \\ \amp = \ccmint{a_1}{a_2}{f(u)}{u} + A_2(t) \text{.} \end{align*}
So, in fact, we can say that
\begin{equation*} A_1(t) = A_2(t) + C \text{,} \end{equation*}
where the shift term is
\begin{equation*} C = \ccmint{a_1}{a_2}{f(u)}{u} \text{.} \end{equation*}
Graph demonstrating the vertical shift between two accumulation functions as a fixed accumulation amount.
Figure 13.4.8. Realizing the vertical shift between two accumulation functions as a fixed accumulation amount.

Example 13.4.9. Two antiderivatives of the cosine function.

The cosine function \(\cos(t)\) is continuous everywhere, so we can create accumulation antiderivatives based at any point. Here are two:
\begin{align*} A_0(t) \amp = \ccmint{0}{t}{\cos(u)}{u} \amp A_{-\pi/2}(t) \amp = \ccmint{-\pi/2}{t}{\cos(u)}{u}\text{.} \end{align*}
We know from Pattern 10.5.2 that
\begin{equation*} A_0(t) = \sin(t) \text{.} \end{equation*}
We could also compute
\begin{equation*} \ccmint{-\pi/2}{0}{\cos(u)}{u} = 1 \text{.} \end{equation*}
(Using the symmetry of the cosine graph, this is one-half of the accumulation we calculated in Example 13.1.7.) So
\begin{equation*} A_{-\pi/2}(t) = 1 + \sin(t) \text{.} \end{equation*}
Now, as at the beginning of the section, suppose we start with a quantity function \(q(t)\) and use Pattern 6.4.7 to create a quantity function
\begin{equation*} s(t) = s_0 + \ccmint{t_0}{t}{r(u)}{u} \end{equation*}
from the rate function \(r(t) = q'(t)\text{.}\) Then we end up with \(s'(t) = q'(t)\text{,}\) and Fact 13.4.6 says that \(s\) must be a vertical shift of \(q\text{.}\) This is almost back where we started — differentiating and then integrating that derivative returns us back to a vertical shift of our original function. However, in creating \(s(t)\text{,}\) we can choose \(s_0\) to be any value we want, so in fact it is possible to have the differentiation-integration cycle come full circle.

Section 13.5 Indefinite integrals

In light of Fact 13.4.4, we borrow integral notation to represent antiderivative possibilities.

Definition 13.5.1. Indefinite integral.

For function \(f(t)\text{,}\) we write
\begin{equation*} \ccmiint{f(t)}{t} \end{equation*}
to represent the family of all possible antiderivatives of \(f(t)\text{.}\)

Warning 13.5.2.

An indefinite integral
\begin{equation*} \ccmiint{f(t)}{t} \end{equation*}
is not a function. You cannot substitute a value \(t = c\) into it.
Fact 13.4.6 tells us that if we know (or can guess) one antiderivative, then we know them all.

Example 13.5.3. The indefinite integral of the cosine function.

How can we describe the family of antiderivatives represented by
\begin{equation*} \ccmiint{\cos(t)}{t} \text{?} \end{equation*}
As in Example 13.4.9, one antiderivative is
\begin{equation*} A_0(t) = \ccmint{0}{t}{\cos(u)}{u} = \sin(t) \text{,} \end{equation*}
and every other antiderivative is a vertical shift of \(A_0(t)\text{.}\) So we can express this family of antiderivatives using a parameter \(C\) to represent that vertical shift:
\begin{equation*} \ccmiint{\cos(t)} = \sin(t) + C \text{.} \end{equation*}

Example 13.5.4. The indefinite integral of the sine function.

For
\begin{equation*} \ccmiint{\sin(t)}{t} \text{,} \end{equation*}
we can follow the same pattern as Example 13.5.3, first creating the accumulation antiderivative
\begin{equation*} A_0(t) = \ccmint{0}{t}{\sin(t)}{t} \text{.} \end{equation*}
From Pattern 10.5.1, we have
\begin{equation*} A_0(t) = 1 - \cos(t) \text{.} \end{equation*}
It is more convenient to take a specific vertical shift of this antiderivative: let
\begin{equation*} F(t) = A_0(t) - 1 = - \cos(t) \text{.} \end{equation*}
This is still an antiderivative of \(\sin(t)\text{,}\) as we know from Pattern 12.6.7 that
\begin{align*} \opddt \bbrac{- \cos(t)} \amp = - \opddt \bbrac{\cos(t)} \\ \amp = - \bbrac{- \sin(t)} \\ \amp = \sin(t) \text{.} \end{align*}
So we can use all vertical shifts of this particular antiderivative to represent the family of antiderivatives:
\begin{equation*} \ccmiint{\sin(t)}{t} = - \cos(t) + C \text{.} \end{equation*}
Here is a small table of indefinite integrals. You should compare with the accumulation function formulas we have encountered in Chapter 7–10, and note where we have vertically shifted those formulas to more simply express the family of antiderivatives, similarly to Example 13.5.4. You should also compare with the differentiation formulas from Chapter 12, as the derivative of the formula for each indefinite integral (with \(dC/dt = 0\text{,}\) always) should equal to the integrand function in the indefinite integral.
Table 13.5.5. Table of integrals
\(\displaystyle \ccmiint{t^m}{t} = \frac{t^{m + 1}}{m + 1} + C\) \((m \neq -1)\) \(\displaystyle \ccmiint{\frac{1}{t}}{t} = \ln\abs{t} + C \)
\(\displaystyle \ccmiint{\cos(t)}{t} = \sin(t) + C \) \(\displaystyle \ccmiint{\sin(t)}{t} = -\cos(t) + C \)
\(\displaystyle \ccmiint{e^t}{t} = e^t + C \)

Section 13.6 Using antiderivatives to compute definite integrals

Example 13.6.1.

Consider
\begin{equation*} \ccmint{3}{7}{e^t}{t} \text{.} \end{equation*}
This integral represents an accumulation over time domain \(3 \le t \le 7\) for a quantity that varies according to rate function \(r(t) = e^t\text{,}\) so we can use our accumulation formula
\begin{equation*} \ccmint{0}{t}{e^u}{u} = e^t - 1 \end{equation*}
(Pattern 9.6.1) to first compute the accumulation over \(0 \le t \le 7\text{,}\) and then subtract out the accumulation over \(0 \le t \le 3\) that we have included in that first calculation but shouldn’t have:
\begin{align*} \ccmint{3}{7}{e^t}{t} \amp = \ccmint{0}{7}{e^t}{t} - \ccmint{0}{3}{e^t}{t} \\ \amp = (e^7 - 1) - (e^3 - 1) \\ \amp = e^7 - e^3 \text{.} \end{align*}
Notice how the minus-one part of the accumulation formula, which we can think of as a vertical shift, cancelled out in the final calculation.
The pattern of Example 13.6.1 works in general. Suppose \(f(t)\) is a continuous function and we want to calculate
\begin{equation*} \ccmint{a}{b}{f(t)}{t} \text{.} \end{equation*}
If we already have a formula for the accumulation function
\begin{equation*} A(t) = \ccmint{t_0}{t}{f(u)}{u} \text{,} \end{equation*}
then we use the integration pattern
\begin{equation*} \ccmint{t_0}{a}{f(u)}{u} + \ccmint{a}{b}{f(u)}{u} = \ccmint{t_0}{b}{f(u)}{u} \end{equation*}
(Pattern 6.3.11) to calculate
\begin{align*} \ccmint{a}{b}{f(u)}{u} \amp = \ccmint{t_0}{b}{f(u)}{u} - \ccmint{t_0}{a}{e^t}{t} \\ \amp = A(b) - A(a) \text{.} \end{align*}
We know that the accumulation function \(A(t)\) is an antiderivative for \(f(t)\text{,}\) so if \(F(t)\) is another antiderivative function it must be a vertical shift of \(A(t)\text{:}\)
\begin{equation*} F(t) = A(t) + C \qquad\implies\qquad A(t) = F(t) - C \text{.} \end{equation*}
So now
\begin{align*} \ccmint{a}{b}{f(u)}{u} \amp = A(b) - A(a) \\ \amp = \bbrac{F(b) - C} - \bbrac{F(a) - C} \\ \amp = F(b) - F(a) \text{.} \end{align*}

Example 13.6.3. Definite integral of a power function.

For
\begin{equation*} \ccmint{4}{9}{\frac{1}{\sqrt{t}}}{t} \text{,} \end{equation*}
the integrand function is
\begin{equation*} f(t) = \frac{1}{\sqrt{t}} = t^{-1/2} \text{.} \end{equation*}
Using Table 13.5.5, we choose antiderivative
\begin{equation*} F(t) = \frac{t^{1/2}}{1/2} = 2 \sqrt{t} \text{.} \end{equation*}
Then
\begin{align*} \ccmint{4}{9}{\frac{1}{\sqrt{t}}}{t} \amp = 2 \sqrt{9} - 2 \sqrt{4} \\ \amp = 2 \text{.} \end{align*}

Example 13.6.4. Definite integral of the reciprocal function.

For
\begin{equation*} \ccmint{-5}{-3}{\frac{1}{t}}{t} \text{,} \end{equation*}
the integrand function is
\begin{equation*} f(t) = \frac{1}{t} \text{.} \end{equation*}
Normally we would use the natural logarithm function to compute a definite integral involving this rate function (see Definition 8.1.1), but this integral is over a negative domain. Instead we use the expanded logarithm function \(F(t) = \ln\abs{t}\) as an antiderivative, so
\begin{align*} \ccmint{-5}{-3}{\frac{1}{t}}{t} \amp = \ln\abs{-3} - \ln\abs{-5} \\ \amp = \ln (3) - \ln (5) \text{.} \end{align*}
Symmetric areas under the graph of the reciprocal function.
Figure 13.6.5. Symmetric areas under the graph of the reciprocal function.
The area represented by this definite integral is below the horizontal axis, and so is negatively oriented. Using the symmetry of the graph of the rate function \(f(t) = 1/t\text{,}\) this area should be equal in magnitude to the area represented by the definite integral
\begin{align*} \ccmint{3}{5}{\frac{1}{t}}{t} \amp = \ccmint{1}{5}{\frac{1}{t}}{t} - \ccmint{1}{3}{\frac{1}{t}}{t}\\ \amp = \ln (5) - \ln (3) \text{,} \end{align*}
as the initial computation confirms.

Example 13.6.6. A definite integral of the sine function.

For
\begin{equation*} \ccmint{\pi/4}{\pi/3}{\sin(t)}{t} \text{,} \end{equation*}
the integrand function is \(\sin(t)\text{.}\) Using Table 13.5.5, we choose antiderivative
\begin{equation*} F(t) = - \cos(t) \text{.} \end{equation*}
Then
\begin{align*} \ccmint{\pi/4}{\pi/3}{\sin(t)}{t} \amp = \bbrac{- \cos(\pi/3)} - \bbrac{- \cos(\pi/4)}\\ \amp = \left(- \frac{1}{2}\right) - \left(- \frac{\sqrt{2}}{2}\right) \\ \amp = \frac{\sqrt{2} - 1}{2} \text{.} \end{align*}

Remark 13.6.7.

With Pattern 13.6.2, we have effectively completed the circle. For quantity function \(q(t)\) over domain \(t_1 \le t \le t_2\text{,}\) we have
\begin{align*} \text{accumulation} \amp = (\text{final amount}) - (\text{initial amount}) \\ \amp = q(t_2) - q(t_1) \text{.} \end{align*}
On the other hand, we now know that the derivative function \(r(t) = q'(t)\) describes the rate of accumulation of \(q(t)\text{,}\) so we expect a definite integral of \(q'(t)\) to also represent an accumulation. Pattern 13.6.2 confirms this, since \(q(t)\) is obviously an antiderivative of its own derivative, so
\begin{equation*} \ccmint{t_1}{t_2}{q'(t)}{t} = q(t_2) - q(t_1) = \text{accumulation} \text{.} \end{equation*}
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