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Chapter 14 Average value and the Fundamental Theorem

Section 14.1 Average rate revisited

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 \(r(t) \text{.}\) Rearranging it to
\begin{equation*} \ccmint{t_1}{t_2}{r(t)}{t} = \avg{r} \cdot \change{t} \text{,} \end{equation*}
where on the left we have an accumulation calculation for a varying rate function (Pattern 9.4.2) and on the right we have an accumulation calcluation for a constant rate (Pattern 1.4.7), leads to the following reinterpretation of average rate.
Interpreting an average rate graphically as an equivalent constant rate.
Figure 14.1.3. Average rate of accumulation as a an equivalent constant rate the rate graph.
In the area bounded between the graph and the horizontal axis in Figure 14.1.3, imagine removing an area equivalent to the negatively-oriented area at the lower-right of the diagram from the top of the positively-oriented area in the upper-left of the diagram. In the area bounded between the horizontal line at the level labelled \(\avg{r} \) and the horizontal axis, imagine stacking the right “half” of this area on top of the left “half”. In this way you can visualize how the net oriented area between graph and horizontal axis is equivalent to the total rectangular area below the \(\avg{r} \) value.

Example 14.1.4. Average speed.

Consider again our Camrose-Wetaskiwin trip from Example 8.2.9, also discussed in Subsection 9.1.1. The actual total distance travelled according to the “more realistic” rate model in Figure 9.1.1 is approximately 50.5 km, and the trip lasts a total of 39 min, for an average speed of
\begin{equation*} \frac{50.5}{39/60} \approx 77.7 \end{equation*}
in kilometres per hour. If it were possible to go from parked to full speed instantaneously, maintain that full speed constantly over the full 39 min, and then go from full speed to parked again instantaneously at the destination, then the exact same distance of 50.5 km would be traversed over such a trip. In Figure 14.1.5, this equality of distances is represented by equality of the red and blue areas.

Aside: Terminology.

Graph comparing accumulation as area between constant average rate and variable rate in a rate-of-travel model.
Figure 14.1.5. Maintaining a constant speed at the average speed value traverses the same distance.

Section 14.2 Average value of a function

What if we tried to view a quantity function as somehow being a rate function, and then computed the “average rate”?

Example 14.2.1. 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 14.2.2. 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 10.4.12 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 14.2.3. Average temperature over a twenty-four-hour period.
In light of Pattern 14.1.1, we expect the area under the temperature graph to equal the rectangular area under the constant average temperature value over the same time period.
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 14.2.4. Average temperature over a twenty-four-hour period achieves the same total “accumulation” of degree-hours as measured by the integral.

Remark 14.2.5.

In Example 14.2.1, 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 14.2.6. 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*}

Example 14.2.7. 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 14.2.6, 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 14.2.8. Computing average temperature from temperature samples.

Let’s return to the situation of Example 14.2.1, 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 14.2.6.

Remark 14.2.9.

The analysis strategy in Example 14.2.8 is a powerful method in engineering and the sciences: if can you make an approximation of some value in such a way that the pattern of your approximation can be interpreted as a Riemann sum, then you can calculate that value exactly with an integral.

Section 14.3 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 14.3.2. 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 14.3.3. A continuous function must achieve its average value.

Example 14.3.4. 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 14.3.5. 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 14.3.6. A parabola achieving its average value twice.

Section 14.4 The Fundamental Theorem

Let’s review.
We have learned how to use average rates of accumulation to approximate an instantaneous rate of accumulation 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 instantaneous rate of accumulation of the quantity function at that instant. We then collect those derivative values instant-by-instant into a function called the derivative function \(q'(t) \) for the quantity function.
We have also learned how to use a Riemann sum to approximate the accumulation of some quantity from the rate function \(r(t) \) that models the rate at which the quantity accumulates. 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 9.4.7, we can combine an accumulation function with an initial value to recover the quantity function \(q(t) \text{.}\)
We would expect that performing both processes one after the other should return us to where we started:
  • If we begin with a quantity function \(q(t) \text{,}\) compute the rate at which the quantity accumulates as represented by the derivative function \(q'(t) \text{,}\) use that derivative function as the rate function in an integral based at some initial time \(t = t_0 \) to create an accumulation function, and then combine that accumulation function with the initial value \(q(t_0) \text{,}\) we should be back at the original quantity function.
  • 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. We verified that it works for power functions in Section 10.5 (up to knowing the initial value). We assumed that it works to help us analyze derivatives and integrals of logarithms and exponentials.
However, if we discard the underlying ideas of varying quantities and instantaneous rates and accumulation and only deal with the mathematical concept of functions, then mathematically there is no reason to think that either of the differentiate-then-integrate or integrate-then-differentiate patterns described above should work, as the two processes involved have seemingly nothing to do with each other: one process uses secant-line slopes to approximate a tangent-line slope, while the other uses rectangles to approximate an area. Why would performing one process after the other return you to where you started? But, in fact, our intuition based on quantities, rates, and accumulation is correct.

Justification.

As usual, to compute the derivative of \(A(t) \text{,}\) we consider a forward/backward difference quotient \(\inlineslope{A}{t} \text{.}\) In the numerator, first consider
\begin{equation*} A(t + \change{t}) = \ccmint{a}{t + \change{t}}{f(u)}{u} \text{.} \end{equation*}
Regardless of whether \(\change{t} \) is positive or negative, assuming that it is small enough that \(t + \change{t} \) remains within the domain \(a \le t \le b \text{,}\) we can use Property 5 of Pattern 9.3.11 to split the integral:
\begin{equation*} A(t + \change{t}) = \ccmint{a}{t}{f(u)}{u} + \ccmint{t}{t + \change{t}}{f(u)}{u} \text{.} \end{equation*}
But now that first integral is precisely \(A(t) \text{,}\) so in the subtraction of \(\change{A} \) we will have cancellation:
\begin{align*} \change{A} \amp = A(t + dt) - A(t) \\ \amp = \left( A(t) + \ccmint{t}{t + \change{t}}{f(u)}{u} \right) - A(t) \\ \amp = \ccmint{t}{t + \change{t}}{f(u)}{u} \text{.} \end{align*}
We now have a simplified version of the numerator \(\change{A} \) to use in our average rate difference quotient:
\begin{equation*} \slope{A}{t} = \frac{A(t + \change{t}) - A(t)}{\change{t}} = \frac{1}{\change{t}} \ccmint{t}{t + \change{t}}{f(u)}{u}\text{.} \end{equation*}
This looks like the formula for the average value of a function! (See Definition 14.2.6.) Indeed, the integral is over the domain \(t \le u \le t + \change{t} \text{,}\) and \(\change{t} \) is the duration of that domain. So
\begin{equation*} \slope{A}{t} = \avg{f} \end{equation*}
over domain \(t \le u \le t + \change{t} \text{.}\) The Mean Value Theorem says that there is at least one time \(c \) in that domain where \(f(c) = \avg{f} \text{.}\) But when \(\change{t} \approx 0 \text{,}\) then
\begin{equation*} t + \change{t} \approx t \text{,} \end{equation*}
and so this \(c \) that is between \(t \) and \(t + \change{t} \) must be “squeezed” down to \(c \approx t \) as well. Along with it, we have \(f(c) \approx f(t) \) because \(f \) is continuous. Ultimately,
\begin{equation*} \slope{A}{t} = \avg{f} = f(c) \to f(t) \text{,} \end{equation*}
so that \(dA/dt = f(t) \text{,}\) as desired.
So the accumulation function \(A(t) \) is differentiable, and Pattern 5.7.11 says that a differentiable 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 14.4.2.

The Fundamental Theorem verifies that integrate-then-differentiate returns back to the original function. Even more, it strengthens the interpretation of the derivative function as measuring instantaneous rate of accumulation. If \(q(t) \) is a quantity function and \(r(t) \) is a model for the rate at which \(q \) accumulates, then from \(r(t) \) we can recover the quantity function using Pattern 9.4.7:
\begin{equation*} q(t) = q(a) + \ccmint{a}{t}{r(u)}{u} \text{.} \end{equation*}
Then, using the Fundamental Theorem and the properties of differentiation from Section 6.3, 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*}
so that the derivative function matches the initial model for the rate at which \(q \) accumulates.

Example 14.4.3. Confirming the Fundamental Theorem in a simple case.

The function \(f(t) = t \) is obviously continuous. Using simple geometry (see Example 10.4.4) 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 Power Rule 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*}
As predicted by the Fundamental Theorem, the derivative of the accumulation function is just the original function used to create it.

Checkpoint 14.4.4. Irrelevance of “base” point in the Fundamental Theorem.

Repeat the calculations in Example 14.4.3, again for the function \(f(t) = t \text{,}\) but this time using the accumulation function
\begin{equation*} A_1(t) = \ccmint{1}{t}{f(u)}{u} \text{.} \end{equation*}
Once again, begin by computing a formula in \(t \) for the integral in this definition just by the geometry of the area to which it corresponds. Then differentiate that formula.

Example 14.4.5. Derivative of \(\ln(t) \) and \(\ln\abs{t} \).

We can now use the Fundamental Theorem to verify our formula for the derivative of the natural logarithm function (Pattern 11.3.1):
\begin{equation*} \ddt{\bbrac{\ln (t)}} = \opddt \ccmint{1}{t}{\frac{1}{u}}{u} = \frac{1}{t}\text{.} \end{equation*}
We can similarly verify Pattern 11.3.6 using our interpretation of \(\ln\abs{t} \) at the end of Subsection 11.3.2:
\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 \text{.}\end{cases} \end{equation*}
From this,
\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 14.4.6. Integrals of sine and cosine.

In Section 13.6, we used the example of a spring attached to a mass as inspiration for integration formulas for sine and cosine.
\begin{align*} \ccmint{0}{t}{\sin(u)}{u} \amp = 1 - \cos(t) \amp \ccmint{0}{t}{\cos(u)}{u} \amp = \sin(t) \end{align*}
(See Pattern 13.6.1 and Pattern 13.6.2.)
Starting with the formula for the integral of sine, let’s differentiate both sides, where on the left we may now apply the Fundamental Theorem to be sure of the result, and on the right we have the confidence of Pattern 13.5.8, with the cosine derivative formula verified in Checkpoint 13.5.9.
\begin{align*} \opddt \, \ccmint{0}{t}{\sin(u)}{u} \amp = \sin(t) \amp \opddt \bbrac{1 - \cos(t)} \amp = \ddt{1} - \ddt{\bbrac{\cos(t)}}\\ \amp \amp \amp = 0 - \bbrac{- \sin(t)} \\ \amp \amp \amp = \sin(t) \end{align*}
We got the same result both times! While this doesn’t fully verify our integral formula for sine (we’ll do that later this chapter), it gives confidence that the formula has the right “form”.
Similarly, separately differentiating each side of our integral formula for cosine gives the same result in both calculationsx.
\begin{align*} \opddt \, \ccmint{0}{t}{\cos(u)}{u} \amp = \cos(t) \amp \ddt{\bbrac{\sin(t)}} \amp = \cos(t) \end{align*}

Section 14.5 Antiderivatives

The Fundamental Theorem of Calculus tells that carrying out an integrate-then-differentiate cycle will return us back to the original function. This even works if we insert the addition of an initial value into the process, as in Pattern 9.4.7. If \(q \) is a quantity that is varying over time and we have a model rate function \(r(t) \) that describes the rate at which \(q \) accumulates, we can use a definite integral to recover a model for the quantity function
\begin{equation*} q(t) = q_0 + \ccmint{t_0}{t}{r(u)}{u} \text{,} \end{equation*}
where \(q_0 \) represents a chosen or observed “initial value” \(q(t_0) = q_0 \text{.}\) If we then differentiate this quantity function, we end up back at the rate function:
\begin{equation*} q'(t) = r(t) \text{.} \end{equation*}
(See the calculation in Remark 14.4.2.)
What about a differentiate-then-integrate cycle beginning with a model for the quantity function \(q(t) \text{?}\) Checkpoint 10.5.1 demonstrates that we won’t always necessarily end up back at the original \(q(t) \text{.}\) Let’s investigate why more generally. Since the derivative function \(q'(t) \) measures instantaneous rate of accumulation, we can use it as a rate model for how \(q \) varies: \(r(t) = q'(t) \text{.}\) We can then use this rate function to create a quantity function \(s(t) \) through the following steps. (We are choosing to represent this “recovered” quantity function by \(s(t) \) instead of \(q(t) \) so as to not bias ourselves into assuming that the cycle will return us to the original quantity function.) First we create an accumulation function based at some choice of “initial” time \(t_0 \text{:}\)
\begin{equation*} A(t) = \ccmint{t_0}{t}{ r(u) }{u} \text{.} \end{equation*}
Then we combine that with some choice of “initial” quantity \(s_0 \text{:}\)
\begin{equation*} s(t) = s_0 + \ccmint{t_0}{t}{r(u)}{u} \text{.} \end{equation*}
Note that
\begin{equation*} s(t_0) = s_0 + \ccmint{t_0}{t_0}{r(u)}{u} = s_0 + 0 = s_0 \text{.} \end{equation*}
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 14.4.2, differentiating the formula for \(s(t) \) above arrives at
\begin{equation*} s'(t) = r(t) = q'(t) \text{.} \end{equation*}

Definition 14.5.1. 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 14.5.2.

For \(f(t) \) and \(F(t) \) in Definition 14.5.1, 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 14.5.3. Two antiderivatives for the same function.

Consider
\begin{align*} F_1(t) \amp = t^2 \text{,} \amp F_2(t) \amp = t^2 + 1 \text{.} \end{align*}
Using the Power Rule and the general differentiation patterns in Chapter 6, we have both
\begin{align*} F_1'(t) \amp = 2 t \text{,} \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 14.5.5. Antiderivative of a linear function.

In Example 14.4.3, beginning with the basic linear function \(f(t) = t \) we:
  1. Formed the accumulation function
    \begin{equation*} A_0(t) = \ccmint{0}{t}{f(u)}{u} \text{.} \end{equation*}
  2. Used geometry to compute the integral defining the accumulation function to obtain an explicit formula in \(t \) for it.
  3. Differentiated that formula for the accumulation function.
At the end of these steps, the result was
\begin{equation*} \ddt{A_0} = f(t) \text{,} \end{equation*}
confirming the Fundamental Theorem in that case. But in the language of Definition 14.5.1, another point of view on these calculations is that \(A_0(t) \) became an antiderivative of \(f(t) \).
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 14.5.3 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 measures instantaneous rates, 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 14.5.6 with \(q_1 \) in place of \(F_1 \) and \(q_2 \) in place of \(F_2 \text{.}\)
We are now in a position to verify our formulas for the integrals of the natural exponential function and of the sine and cosine functions.

Example 14.5.8. Two antiderivatives of the natural exponential function.

The Fundamental Theorem tells us that the accumulation function
\begin{equation*} \ccmint{0}{t}{ e^u }{u} \end{equation*}
is an antiderivative for the natural exponential function \(\exp(t) \text{.}\) Using Pattern 12.6.2 and the general differentiation rules from Chapter 6, it is straightforward to verify that the expression
\begin{equation*} e^t - 1 \end{equation*}
also forms an antiderivative function for \(\exp(t) \text{.}\) Fact 14.5.6 now implies that these two functions must be vertical shifts of one another; that is, there is some value of the constant \(C \) for which
\begin{equation*} \ccmint{0}{t}{ e^u }{u} = e^t - 1 + C \text{.} \end{equation*}
Two functions that are equal must always produce the same output when fed the same input. In the case of \(t = 0 \text{,}\) this means
\begin{equation*} \ccmint{0}{0}{ e^u }{u} = e^0 - 1 + C = 1 - 1 + C = C \text{.} \end{equation*}
But the integral on the left is accumulation over a time period of duration \(0 \text{,}\) and so that accumulation is \(0 \) (Property 1 of Convention 9.3.12). This means that \(C = 0 \text{,}\) and so in fact we have
\begin{equation*} \ccmint{0}{t}{ e^u }{u} = e^t - 1 \text{,} \end{equation*}
confirming the formula provided in Pattern 12.7.1.
How does Fact 14.5.6 apply to the case of two different accumulation functions for the same rate function? If the function \(f(t) \) is continuous on a large domain, when creating an accumulation antiderivative as in Fact 14.5.4, 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} \text{,} \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 14.5.10. Realizing the vertical shift between two accumulation functions as a fixed accumulation amount.

Example 14.5.11. 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 13.6.2 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 14.2.7.) 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 9.4.7 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 14.5.6 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 while the differentiate-then-integrate cycle won’t necessarily come full circle back to the original function we started with, careful choice of an additional “initial value” can always be made to ensure it does.

Section 14.6 Indefinite integrals

In light of Fact 14.5.4, we borrow integral notation to represent antiderivative possibilities.

Definition 14.6.1. 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 14.6.2.

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 14.5.6 tells us that if we know (or can guess) one antiderivative, then we know them all.

Example 14.6.3. 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 14.5.11, 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 14.6.4. 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 14.6.3, first creating the accumulation antiderivative
\begin{equation*} A_0(t) = \ccmint{0}{t}{\sin(t)}{t} \text{.} \end{equation*}
From Pattern 13.6.1, 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 13.5.8 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 10–13, and note where we have vertically shifted those formulas to more simply express the family of antiderivatives, similarly to Example 14.6.4. You should also compare with the differentiation formulas from Chapter 6, 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 14.6.5. 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 14.7 Using antiderivatives to compute definite integrals

Example 14.7.1.

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 12.7.1) 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 14.7.1 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 9.3.11) 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 14.7.3. 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 14.6.5, 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 14.7.4. 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 11.1.1), 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 14.7.5. 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 14.7.6. 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 14.6.5, 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 14.7.7.

With Pattern 14.7.2, 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 14.7.2 confirms this, since \(q(t) \) is obviously an antiderivative of its own derivative, so
\begin{equation*} \ccmint{t_1}{t_2}{ \dqdt }{t} = q(t_2) - q(t_1) = \text{accumulation} \text{.} \end{equation*}
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