Active Calculus

Subsection 4.4.1 The Fundamental Theorem of Calculus

Suppose we know the position function \(s(t)\) and the velocity function \(v(t)\) of an object moving in a straight line, and for the moment let us assume that \(v(t)\) is positive on \([a,b]\text{.}\) Then, as shown in Figure 4.4.2, we know two different ways to compute the distance, \(D\text{,}\) the object travels: one is that \(D = s(b) - s(a)\text{,}\) the object's change in position. The other is the area under the velocity curve, which is given by the definite integral, so \(D = \int_a^b v(t) \, dt\text{.}\)

Since both of these expressions tell us the distance traveled, it follows that they are equal, so

Equation (4.4.1) holds even when velocity is sometimes negative, because \(s(b) - s(a)\text{,}\)the object's change in position, is also measured by the net signed area on \([a,b]\) which is given by \(\int_a^b v(t) \, dt\text{.}\)

Perhaps the most powerful fact Equation (4.4.1) reveals is that we can compute the integral's value if we can find a formula for \(s\text{.}\) Remember, \(s\) and \(v\) are related by the fact that \(v\) is the derivative of \(s\text{,}\) or equivalently that \(s\) is an antiderivative of \(v\text{.}\)

Note the key lesson of Example 4.4.3: to find the distance traveled, we need to compute the area under a curve, which is given by the definite integral. But to evaluate the integral, we can find an antiderivative, \(s\text{,}\) of the velocity function, and then compute the total change in \(s\) on the interval. In particular, we can evaluate the integral without computing the limit of a Riemann sum.

It will be convenient to have a shorthand symbol for a function's antiderivative. For a continuous function \(f\text{,}\) we will often denote an antiderivative of \(f\) by \(F\text{,}\) so that \(F'(x) = f(x)\) for all relevant \(x\text{.}\) Using the notation \(V\) in place of \(s\) (so that \(V\) is an antiderivative of \(v\)) in Equation (4.4.1), we can write

Now, to evaluate the definite integral \(\int_a^b f(x) \, dx\) for an arbitrary continuous function \(f\text{,}\) we could certainly think of \(f\) as representing the velocity of some moving object, and \(x\) as the variable that represents time. But Equations (4.4.1) and (4.4.2) hold for any continuous velocity function, even when \(v\) is sometimes negative. So Equation (4.4.2) offers a shortcut route to evaluating any definite integral, provided that we can find an antiderivative of the integrand. The Fundamental Theorem of Calculus (FTC) summarizes these observations.

A common alternate notation for \(F(b) - F(a)\) is

where we read the righthand side as “the function \(F\) evaluated from \(a\) to \(b\text{.}\)” In this notation, the FTC says that

The FTC opens the door to evaluating a wide range of integrals if we can find an antiderivative \(F\) for the integrand \(f\text{.}\) For instance since \(\frac{d}{dx}[\frac{1}{3}x^3] = x^2\text{,}\) the FTC tells us that

But finding an antiderivative can be far from simple; it is often difficult or even impossible. While we can differentiate just about any function, even some relatively simple functions don't have an elementary antiderivative. A significant portion of integral calculus (which is the main focus of second semester college calculus) is devoted to the problem of finding antiderivatives.

Subsection 4.4.2 Basic antiderivatives

The general problem of finding an antiderivative is difficult. In part, this is due to the fact that we are trying to undo the process of differentiating, and the undoing is much more difficult than the doing. For example, while it is evident that an antiderivative of \(f(x) = \sin(x)\) is \(F(x) = -\cos(x)\) and that an antiderivative of \(g(x) = x^2\) is \(G(x) = \frac{1}{3} x^3\text{,}\) combinations of \(f\) and \(g\) can be far more complicated. Consider the functions

What is involved in trying to find an antiderivative for each? From our experience with derivative rules, we know that derivatives of sums and constant multiples of basic functions are simple to execute, but derivatives involving products, quotients, and composites of familiar functions are more complicated. Therefore, it stands to reason that antidifferentiating products, quotients, and composites of basic functions may be even more challenging. We defer our study of all but the most elementary antiderivatives to later in the text.

We do note that whenever we know the derivative of a function, we have a function-derivative pair, so we also know the antiderivative of a function. For instance, since we know that

we also know that \(F(x) = -\cos(x)\) is an antiderivative of \(f(x) = \sin(x)\text{.}\) \(F\) and \(f\) together form a function-derivative pair. Clearly, every basic derivative rule leads us to such a pair, and thus to a known antiderivative.

In Activity 4.4.3, we will construct a list of the basic antiderivatives we know at this time. Those rules will help us antidifferentiate sums and constant multiples of basic functions. For example, since \(-\cos(x)\) is an antiderivative of \(\sin(x)\) and \(\frac{1}{3}x^3\) is an antiderivative of \(x^2\text{,}\) it follows that

is an antiderivative of \(f(x) = 5\sin(x) - 4x^2\text{,}\) by the sum and constant multiple rules for differentiation.

Finally, before proceeding to build a list of common functions whose antiderivatives we know, we recall that each function has more than one antiderivative. Because the derivative of any constant is zero, we may add a constant of our choice to any antiderivative. For instance, we know that \(G(x) = \frac{1}{3}x^3\) is an antiderivative of \(g(x) = x^2\text{.}\) But we could also have chosen \(G(x) = \frac{1}{3}x^3 + 7\text{,}\) since in this case as well, \(G'(x) = x^2\text{.}\) If \(g(x) = x^2\text{,}\) we say that the general antiderivative of \(g\) is

where \(C\) represents an arbitrary real number constant. Regardless of the formula for \(g\text{,}\) including \(+C\) in the formula for its antiderivative \(G\) results in the most general possible antiderivative.

Our current interest in antiderivatives is so that we can evaluate definite integrals by the Fundamental Theorem of Calculus. For that task, the constant \(C\) is irrelevant, and we usually omit it. To see why, consider the definite integral

For the integrand \(g(x) = x^2\text{,}\) suppose we find and use the general antiderivative \(G(x) = \frac{1}{3} x^3 + C\text{.}\) Then, by the FTC,

Observe that the \(C\)-values appear as opposites in the evaluation of the integral and thus do not affect the definite integral's value.

In the following activity, we work to build a list of basic functions whose antiderivatives we already know.

Subsection 4.4.3 The total change theorem

Let us review three interpretations of the definite integral.

• For a moving object with instantaneous velocity \(v(t)\text{,}\) the object's change in position on the time interval \([a,b]\) is given by \(\int_a^b v(t) \, dt\text{,}\) and whenever \(v(t) \ge 0\) on \([a,b]\text{,}\) \(\int_a^b v(t) \, dt\) tells us the total distance traveled by the object on \([a,b]\text{.}\)

• For any continuous function \(f\text{,}\) its definite integral \(\int_a^b f(x) \, dx\) represents the net signed area bounded by \(y = f(x)\) and the \(x\)-axis on \([a,b]\text{,}\) where regions that lie below the \(x\)-axis have a minus sign associated with their area.

• The value of a definite integral is linked to the average value of a function: for a continuous function \(f\) on \([a,b]\text{,}\) its average value \(f_{\operatorname{AVG} [a,b]}\) is given by

  \begin{equation*} f_{\operatorname{AVG} [a,b]} = \frac{1}{b-a} \int_a^b f(x) \, dx\text{.} \end{equation*}

The Fundamental Theorem of Calculus now enables us to evaluate exactly (without taking a limit of Riemann sums) any definite integral for which we are able to find an antiderivative of the integrand.

A slight change in perspective allows us to gain even more insight into the meaning of the definite integral. Recall Equation (4.4.2), where we wrote the Fundamental Theorem of Calculus for a velocity function \(v\) with antiderivative \(V\) as

If we instead replace \(V\) with \(s\) (which represents position) and replace \(v\) with \(s'\) (since velocity is the derivative of position), Equation (4.4.2) then reads as

In words, this version of the FTC tells us that the total change in an object's position function on a particular interval is given by the definite integral of the position function's derivative over that interval.

Of course, this result is not limited to only the setting of position and velocity. Writing the result in terms of a more general function \(f\text{,}\) we have the Total Change Theorem.

The Total Change Theorem tells us more about the relationship between the graph of a function and that of its derivative. Recall that heights on the graph of the derivative function are equal to slopes on the graph of the function itself. If instead we know \(f'\) and are seeking information about \(f\text{,}\) we can say the following:

differences in heights on \(f\) correspond to net signed areas bounded by \(f'\text{.}\)

To see why this is so, consider the difference \(f(1) - f(0)\text{.}\) This value is 3, because \(f(1) = 3\) and \(f(0) = 0\text{,}\) but also because the net signed area bounded by \(y = f'(x)\) on \([0,1]\) is 3. That is,

In addition to this observation about area, the Total Change Theorem enables us to answer questions about a function whose rate of change we know.

Suppose that pollutants are leaking out of an underground storage tank at a rate of \(r(t)\) gallons/day, where \(t\) is measured in days. It is conjectured that \(r(t)\) is given by the formula \(r(t) = 0.0069t^3 -0.125t^2+11.079\) over a certain 12-day period. The graph of \(y=r(t)\) is given in Figure 4.4.8. What is the meaning of \(\int_4^{10} r(t) \, dt\) and what is its value? What is the average rate at which pollutants are leaving the tank on the time interval \(4 \le t \le 10\text{?}\)

Solution

Since \(r(t) \ge 0\text{,}\) the value of \(\int_4^{10} r(t) \, dt\) is the area under the curve on the interval \([4,10]\text{.}\) A Riemann sum for this area will have rectangles with heights measured in gallons per day and widths measured in days, so the area of each rectangle will have units of

\begin{equation*} \frac{\text{gallons} }{\text{day} } \cdot \text{days} = \text{gallons}\text{.} \end{equation*}

Thus, the definite integral tells us the total number of gallons of pollutant that leak from the tank from day 4 to day 10. The Total Change Theorem tells us the same thing: if we let \(R(t)\) denote the total number of gallons of pollutant that have leaked from the tank up to day \(t\text{,}\) then \(R'(t) = r(t)\text{,}\) and

\begin{equation*} \int_4^{10} r(t) \, dt = R(10) - R(4)\text{,} \end{equation*}

the number of gallons that have leaked from day 4 to day 10.

To compute the exact value of the integral, we use the Fundamental Theorem of Calculus. Antidifferentiating \(r(t) = 0.0069t^3 -0.125t^2+11.079\text{,}\) we find that

\begin{align*} \int_4^{10} 0.0069t^3 -0.125t^2+11.079 \, dt =\mathstrut \amp \left. 0.0069 \cdot \frac{1}{4} \, t^4 - 0.125 \cdot \frac{1}{3} t^3 + 11.079t \right|_4^{10}\\ \approx\mathstrut \amp 44.282\text{.} \end{align*}

Thus, approximately 44.282 gallons of pollutant leaked over the six day time period.

To find the average rate at which pollutant leaked from the tank over \(4 \le t \le 10\text{,}\) we compute the average value of \(r\) on \([4,10]\text{.}\) Thus,

\begin{equation*} r_{\operatorname{AVG} [4,10]} = \frac{1}{10-4} \int_4^{10} r(t) \, dt \approx \frac{44.282}{6} = 7.380 \end{equation*}

gallons per day.

Activity 4.4.4.

During a 40-minute workout, a person riding an exercise machine burns calories at a rate of \(c\) calories per minute, where the function \(y = c(t)\) is given in Figure 4.4.9. On the interval \(0 \le t \le 10\text{,}\) the formula for \(c\) is \(c(t) = -0.05t^2 + t + 10\text{,}\) while on \(30 \le t \le 40\text{,}\) its formula is \(c(t) = -0.05t^2 + 3t - 30\text{.}\)

1. What is the exact total number of calories the person burns during the first 10 minutes of her workout?

2. Let \(C(t)\) be an antiderivative of \(c(t)\text{.}\) What is the meaning of \(C(40) - C(0)\) in the context of the person exercising? Include units on your answer.

3. Determine the exact average rate at which the person burned calories during the 40-minute workout.

4. At what time(s), if any, is the instantaneous rate at which the person is burning calories equal to the average rate at which she burns calories, on the time interval \(0 \le t \le 40\text{?}\)

Hint

1. What are the units on the area of a rectangle found in a Riemann sum for the function \(y= c(t)\text{?}\)

2. Use the FTC.

3. Recall the formula for \(c_{\operatorname{AVG} [0,40]}\text{.}\)

4. Think carefully about which function tells you the instantaneous rate at which calories are burned.

Answer

1. The person burned approximately 133.33 calories in the first 10 minutes of the workout.

2. \(C(40) - C(0) = \int_0^{40} C'(t) \, dt = \int_0^{40} c(t) \, dt\) is the total calories burned on \([0,40]\text{.}\)

3. The exact average rate at which the person burned calories on \(0 \le t \le 40\) is

   \begin{equation*} c_{\operatorname{AVG} [0,40]} = \frac{1}{40-0} \int_0^{40} c(t) \, dt = \frac{1}{40} \cdot \frac{1700}{3} = \frac{1700}{120} \approx 14.17 \ \text{cal/min}\text{.} \end{equation*}

4. One time at which the instantaneous rate at which calories are burned equals the average rate on \([0,40]\) is \(t = \frac{5}{3}(6 - \sqrt{6}) \approx 5.918\text{.}\)

Solution

1. Since the units on a rectangle in a Riemann sum are cal/min for the height and min for the width, the units on the area of such a rectangle are calories, and hence the units on area under the curve \(y = c(t)\) are given in total calories. Hence, the total calories burned during the first 10 minutes of the workout is given by the definite integral \(\int_0^{10} c(t) \, dt\text{.}\) We use the FTC and evaluate the integral, finding that

   \begin{align*} \int_0^{10} (-0.05t^2 + t + 10) \, dt =\mathstrut \amp \left. \left( -\frac{0.05}{3} t^3 + \frac{1}{2} t^2 + 10t \right) \right|_0^{10}\\ =\mathstrut \amp \left( -\frac{0.05}{3} (10)^3 + \frac{1}{2} (10)^2 + 10(10) \right) - (-0 + 0 + 0)\\ =\mathstrut \amp \frac{400}{3} \ \text{calories}\text{.} \end{align*}

   Thus, the person burned approximately 133.33 calories in the first 10 minutes of the workout.

2. We observe first that by the Total Change Theorem, \(C(40) - C(0) = \int_0^{40} C'(t) \, dt = \int_0^{40} c(t) \, dt\text{,}\) and therefore, as discussed in (a), the meaning of this value is the total calories burned on \([0,40]\text{.}\)

3. The exact average rate at which the person burned calories on \(0 \le t \le 40\) is given by

   \begin{equation*} c_{\text{AVG} [0,40]} = \frac{1}{40-0} \int_0^{40} c(t) \, dt\text{.} \end{equation*}

   To calculate \(\int_0^{40} c(t) \, dt\text{,}\) we recognize that \(c(t)\) is defined in piecewise fashion, and use the additive property of the definite integral, which tells us that

   \begin{equation*} \int_0^{40} c(t) \, dt = \int_0^{10} c(t) \, dt + \int_{10}^{30} c(t) \, dt + \int_{30}^{40} c(t) \, dt\text{.} \end{equation*}

   We know from our work in (a) that \(\int_0^{10} c(t) \, dt = \frac{400}{3}\text{.}\) Since \(c(t) = 15\) is constant on \(10 \le t \le 20\text{,}\) it follows that \(\int_{10}^{30} c(t) \, dt = 15 \cdot 30 = 300\text{.}\) And finally, it is straightforward to show using \(c(t) = -0.05t^2 + 3t - 30\) (or symmetry) on \(30 \le t \le 40\) that \(\int_{30}^{40} c(t) \, dt = \frac{400}{3}\text{.}\) Hence,

   \begin{align*} \int_0^{40} c(t) \, dt =\mathstrut \amp \int_0^{10} c(t) \, dt + \int_{10}^{30} c(t) \, dt + \int_{30}^{40} c(t) \, dt\\ =\mathstrut \amp \frac{400}{3} + 300 + \frac{400}{3}\\ =\mathstrut \amp \frac{1700}{3} \approx 566.67 \ \text{calories}\text{.} \end{align*}

   Now, it follows that the exact average rate at which calories were burned on \([0,40]\) is

   \begin{equation*} c_{\text{AVG} [0,40]} = \frac{1}{40-0} \int_0^{40} c(t) \, dt = \frac{1}{40} \cdot \frac{1700}{3} = \frac{1700}{120} \approx 14.17 \ \text{cal/min}\text{.} \end{equation*}

4. It makes sense intuitively that there must be at least one time at which the instantaneous rate at which calories are burned equals the average rate at which calories are burned, as it would be impossible for a continuous instantaneous rate of change to always be above its average value. Since we know from (c) that \(c_{\text{AVG} [0,40]} = \frac{85}{6}\text{,}\) and \(c(t)\) tells us the instantaneous rate at which calories are burned, it follows that we want to solve the equation

   \begin{equation*} c(t) = \frac{85}{6}\text{.} \end{equation*}

   From the graph, it appears that there are two such values of \(t\) for which this equation is true, one in the first ten minutes, and one in the last ten. For instance, solving

   \begin{equation*} -0.05t^2 + t + 10 = \frac{85}{6}\text{,} \end{equation*}

   it follows that \(t = \frac{5}{3}(6 \pm \sqrt{6}) \approx 14.082, 5.918\text{,}\) only the second of which lies in \(0 \le t \le 10\text{.}\) So one time at which the instantaneous rate at which calories are burned equals the average rate on \([0,40]\) is \(t = \frac{5}{3}(6 - \sqrt{6}) \approx 5.918\text{.}\) Similar reasoning leads to the second time that lies in \([30,40]\text{.}\)

Subsection 4.4.4 Summary

• We can find the exact value of a definite integral without taking the limit of a Riemann sum or using a familiar area formula by finding the antiderivative of the integrand, and hence applying the Fundamental Theorem of Calculus.

• The Fundamental Theorem of Calculus says that if \(f\) is a continuous function on \([a,b]\) and \(F\) is an antiderivative of \(f\text{,}\) then

  \begin{equation*} \int_a^b f(x) \, dx = F(b) - F(a)\text{.} \end{equation*}

  Hence, if we can find an antiderivative for the integrand \(f\text{,}\) evaluating the definite integral comes from simply computing the change in \(F\) on \([a,b]\text{.}\)

• A slightly different perspective on the FTC allows us to restate it as the Total Change Theorem, which says that

  \begin{equation*} \int_a^b f'(x) \, dx = f(b) - f(a)\text{,} \end{equation*}

  for any continuously differentiable function \(f\text{.}\) This means that the definite integral of the instantaneous rate of change of a function \(f\) on an interval \([a,b]\) is equal to the total change in the function \(f\) on \([a,b]\text{.}\)