Active Calculus

Subsection 5.1.1 Constructing the graph of an antiderivative

Preview Activity 5.1.1 demonstrates that when we can find the exact area under the graph of a function on any given interval, it is possible to construct a graph of the function's antiderivative. That is, we can find a function whose derivative is given. We can now determine not only the overall shape of the antiderivative graph, but also the actual height of the graph at any point of interest.

This is a consequence of the Fundamental Theorem of Calculus: if we know a function \(f\) and the value of the antiderivative \(F\) at some starting point \(a\text{,}\) we can determine the value of \(F(b)\) via the definite integral. Since \(F(b) - F(a) = \int_a^b f(x) \, dx\text{,}\) it follows that

We can also interpret the equation \(F(b) - F(a) = \int_a^b f(x) \, dx\) in terms of the graphs of \(f\) and \(F\) as follows. On an interval \([a,b]\text{,}\)

differences in heights on the graph of the antiderivative given by \(F(b) - F(a)\) correspond to the net signed area bounded by the original function on the interval \([a,b]\text{,}\) which is given by \(\int_a^b f(x) \, dx\text{.}\)

Activity 5.1.2.

Suppose that the function \(y = f(x)\) is given by the graph shown in Figure 5.1.2, and that the pieces of \(f\) are either portions of lines or portions of circles. In addition, let \(F\) be an antiderivative of \(f\) and say that \(F(0) = -1\text{.}\) Finally, assume that for \(x \le 0\) and \(x \ge 7\text{,}\) \(f(x) = 0\text{.}\)

1. On what interval(s) is \(F\) an increasing function? On what intervals is \(F\) decreasing?

2. On what interval(s) is \(F\) concave up? concave down? neither?

3. At what point(s) does \(F\) have a relative minimum? a relative maximum?

4. Use the given information to determine the exact value of \(F(x)\) for \(x = 1, 2, \ldots, 7\text{.}\) In addition, what are the values of \(F(-1)\) and \(F(8)\text{?}\)

5. Based on your responses to all of the preceding questions, sketch a complete and accurate graph of \(y = F(x)\) on the axes provided, being sure to indicate the behavior of \(F\) for \(x \lt 0\) and \(x \gt 7\text{.}\) Clearly indicate the scale on the vertical and horizontal axes of your graph.

6. What happens if we change one key piece of information: in particular, say that \(G\) is an antiderivative of \(f\) and \(G(0) = 0\text{.}\) How (if at all) would your answers to the preceding questions change? Sketch a graph of \(G\) on the same axes as the graph of \(F\) you constructed in (e).

Hint

1. Consider the sign of \(F' = f\text{.}\)

2. Consider the sign of \(F'' = f'\text{.}\)

3. Where does \(F' = f\) change sign?

4. Recall that \(F(1) = F(0) + \int_0^1 f(t) \, dt\text{.}\)

5. Use the function values found in (d) and the earlier information regarding the shape of \(F\text{.}\)

6. Note that \(G(1) = G(0) + \int_0^1 f(t) \, dt\text{.}\)

Answer

1. \(F\) is increasing on \((0,2)\) and \((5,7)\text{;}\) \(F\) is decreasing on \((2,5)\text{.}\)

2. \(F\) is concave up on \((0,1)\text{,}\) \((4,6)\text{;}\) concave down on \((1,3)\text{,}\) \((6,7)\text{;}\) neither on \((3,4)\text{.}\)

3. A relative maximum at \(x = 2\text{;}\) a relative minimum at \(x = 5\text{.}\)

4. \(F(1) = -\frac{1}{2}\text{;}\) \(F(2) = \frac{\pi}{4} - \frac{1}{2}\text{;}\) \(F(3) = \frac{\pi}{4} - 1\text{;}\) \(F(4) = \frac{\pi}{4}-2\text{;}\) \(F(5) = \frac{\pi}{4} - \frac{5}{2}\text{;}\) \(F(6) = \frac{\pi}{2} - \frac{5}{2}\text{;}\) \(F(7) = \frac{3\pi}{4} - \frac{5}{2}\text{;}\) \(F(8) = \frac{3\pi}{4} - \frac{5}{2}\text{;}\) and \(F(-1) = -1\text{.}\)

5. Use the function values found in (d) and the earlier information regarding the shape of \(F\text{.}\)

6. \(G(x) = F(x) + 1\text{.}\)

Solution

1. Wherever \(F' \gt 0\text{,}\) \(F\) is increasing, so \(F\) is increasing on \((0,2)\) and \((5,7)\text{,}\) while \(F\) is decreasing on \((2,5)\text{.}\)

2. Wherever \(F'' \gt 0\text{,}\) \(F\) is concave up; note particularly that \(F'' \gt 0\) if and only if \(f\) is increasing. Thus, \(F\) is concave up on \((0,1)\text{,}\) \((4,6)\text{,}\) and concave down on \((1,3)\text{,}\) \((6,7)\text{,}\) and neither on \((3,4)\text{.}\)

3. A relative maximum for \(F\) will occur wherever \(F'\) changes from positive to negative, and thus at \(x = 2\text{;}\) similarly, \(F\) has a relative minimum at \(x = 5\text{.}\)

4. Recall that \(F(1) = F(0) + \int_0^1 f(t) \, dt\text{,}\) so \(F(1) = -1 + \frac{1}{2} = -\frac{1}{2}\text{.}\) Similarly, \(F(2) = F(0) + \int_0^1 f(t) \, dt = -1 + \frac{1}{2} + \frac{\pi}{4} = \frac{\pi}{4} - \frac{1}{2}\text{.}\) Continuing these calculations, \(F(3) = \frac{\pi}{4} - 1\text{,}\) \(F(4) = \frac{\pi}{4}-2\text{,}\) \(F(5) = \frac{\pi}{4} - \frac{5}{2}\text{,}\) \(F(6) = \frac{\pi}{2} - \frac{5}{2}\text{,}\) \(F(7) = \frac{3\pi}{4} - \frac{5}{2}\text{.}\) Furthermore, since \(f(t) = 0\) for all \(t \lt 0\) and all \(t \gt 7\text{,}\) it follows \(F(8) = \frac{3\pi}{4} - \frac{5}{2}\) and \(F(-1) = -1\text{.}\)

5. Use the function values found in (d) and the earlier information regarding the shape of \(F\text{.}\)

6. Note that \(G(1) = G(0) + \int_0^1 f(t) \, dt\text{.}\) Since \(G(0) = 0\) (while \(F(0) = -1\)), this changes each response in by 1: \(G(x) = F(x) + 1\text{.}\)

Subsection 5.1.2 Multiple antiderivatives of a single function

In the final question of Activity 5.1.2, we encountered a very important idea: a function \(f\) has more than one antiderivative. Each antiderivative of \(f\) is determined uniquely by its value at a single point. For example, suppose that \(f\) is the function given at left in Figure 5.1.3, and suppose further that \(F\) is an antiderivative of \(f\) that satisfies \(F(0) = 1\text{.}\)

Then, using Equation (5.1.1), we can compute

Similarly, \(F(2) = 1.5\text{,}\) \(F(3) = -0.5\text{,}\) \(F(4) = -2\text{,}\) \(F(5) = -0.5\text{,}\) and \(F(6) = 1\text{.}\) In addition, we can use the fact that \(F' = f\) to ascertain where \(F\) is increasing and decreasing, concave up and concave down, and has relative extremes and inflection points. We ultimately find that the graph of \(F\) is the one given in blue in Figure 5.1.3.

If we want an antiderivative \(G\) for which \(G(0) = 3\text{,}\) then \(G\) will have the exact same shape as \(F\) (since both share the derivative \(f\)), but \(G\) will be shifted vertically from the graph of \(F\text{,}\) as pictured in red in Figure 5.1.3. Note that \(G(1) - G(0) = \int_0^1 f(x) \, dx = 0.5\text{,}\) just as \(F(1) - F(0) = 0.5\text{,}\) but since \(G(0) = 3\text{,}\) \(G(1) = G(0) + 0.5 = 3.5\text{,}\) whereas \(F(1) = 1.5\text{.}\) In the same way, if we assigned a different initial value to the antiderivative, say \(H(0) = -1\text{,}\) we would get still another antiderivative, as shown in magenta in Figure 5.1.3.

This example demonstrates an important fact that holds more generally:

To see why this result holds, observe that if \(G\) and \(H\) are both antiderivatives of \(f\text{,}\) then \(G' = f\) and \(H' = f\text{.}\) Hence,

Since the only way a function can have derivative zero is by being a constant function, it follows that the function \(G - H\) must be constant.

We now see that if a function has at least one antiderivative, it must have infinitely many: we can add any constant of our choice to the antiderivative and get another antiderivative. For this reason, we sometimes refer to the general antiderivative of a function \(f\text{.}\)

To identify a particular antiderivative of \(f\text{,}\) we must know a single value of the antiderivative \(F\) (this value is often called an initial condition). For example, if \(f(x) = x^2\text{,}\) its general antiderivative is \(F(x) = \frac{1}{3}x^3 + C\text{,}\) where we include the “\(+C\)” to indicate that \(F\) includes all of the possible antiderivatives of \(f\text{.}\) If we know that \(F(2) = 3\text{,}\) we substitute 2 for \(x\) in \(F(x) = \frac{1}{3}x^3 + C\text{,}\) and find that

or \(C = 3 - \frac{8}{3} = \frac{1}{3}\text{.}\) Therefore, the particular antiderivative in this case is \(F(x) = \frac{1}{3}x^3 + \frac{1}{3}\text{.}\)

Subsection 5.1.3 Functions defined by integrals

Equation (5.1.1) allows us to compute the value of the antiderivative \(F\) at a point \(b\text{,}\) provided that we know \(F(a)\) and can evaluate the definite integral from \(a\) to \(b\) of \(f\text{.}\) That is,

In several situations, we have used this formula to compute \(F(b)\) for several different values of \(b\text{,}\) and then plotted the points \((b,F(b))\) to help us draw an accurate graph of \(F\text{.}\) This suggests that we may want to think of \(b\text{,}\) the upper limit of integration, as a variable itself. To that end, we introduce the idea of an integral function, a function whose formula involves a definite integral.

Note that because \(x\) is the independent variable in the function \(A\text{,}\) and determines the endpoint of the interval of integration, we need to use a different variable as the variable of integration. A standard choice is \(t\text{,}\) but any variable other than \(x\) is acceptable.

One way to think of the function \(A\) is as the “net signed area from \(a\) up to \(x\)” function, where we consider the region bounded by \(y = f(t)\text{.}\) For example, in Figure 5.1.5, we see a function \(f\) pictured at left, and its corresponding area function (choosing \(a = 0\)), \(A(x) = \int_0^x f(t) \, dt\) shown at right.

The function \(A\) measures the net signed area from \(t = 0\) to \(t = x\) bounded by the curve \(y = f(t)\text{;}\) this value is then reported as the corresponding height on the graph of \(y = A(x)\text{.}\) At http://gvsu.edu/s/cz, we find a java applet¹ that brings the static picture in Figure 5.1.5 to life. There, the user can move the red point on the function \(f\) and see how the corresponding height changes at the light blue point on the graph of \(A\text{.}\)

The choice of \(a\) is somewhat arbitrary. In the activity that follows, we explore how the value of \(a\) affects the graph of the integral function.

Activity 5.1.4.

Suppose that \(g\) is given by the graph at left in Figure 5.1.6 and that \(A\) is the corresponding integral function defined by \(A(x) = \int_1^x g(t) \, dt\text{.}\)

1. On what interval(s) is \(A\) an increasing function? On what intervals is \(A\) decreasing? Why?

2. On what interval(s) do you think \(A\) is concave up? concave down? Why?

3. At what point(s) does \(A\) have a relative minimum? a relative maximum?

4. Use the given information to determine the exact values of \(A(0)\text{,}\) \(A(1)\text{,}\) \(A(2)\text{,}\) \(A(3)\text{,}\) \(A(4)\text{,}\) \(A(5)\text{,}\) and \(A(6)\text{.}\)

5. Based on your responses to all of the preceding questions, sketch a complete and accurate graph of \(y = A(x)\) on the axes provided, being sure to indicate the behavior of \(A\) for \(x \lt 0\) and \(x \gt 6\text{.}\)

6. How does the graph of \(B\) compare to \(A\) if \(B\) is instead defined by \(B(x) = \int_0^x g(t) \, dt\text{?}\)

Hint

1. Where is \(A\) accumulating positive signed area?

2. As \(A\) accumulates positive or negative signed area, where is the rate at which such area is accumulated increasing?

3. Where does \(A\) change from accumulating positive signed area to accumulating negative signed area?

4. Note, for instance, that \(A(2) = \int_1^2 g(t) \, dt\text{.}\)

5. Use your work in (a)-(d) appropriately.

6. What is the value of \(B(0)\text{?}\) How does this compare to \(A(0)\text{?}\)

Answer

1. \(A\) is increasing on \((0,1.5)\text{,}\) \((4,6)\text{;}\) \(A\) is decreasing on \((1.5,4)\text{.}\)

2. \(A\) is concave up on \((0,1)\) and \((3,5)\text{;}\) \(A\) is concave down on \((1,3)\) and \((5,6)\text{.}\)

3. At \(x = 1.5\text{,}\) \(A\) has a relative maximum; \(A\) has a relative minimum at \(x = 4\text{.}\)

4. \(A(0) = -\frac{1}{2}\text{;}\) \(A(1) = 0\text{;}\) \(A(2) = 0\text{;}\) \(A(3) = -2\text{;}\) \(A(4) = -3.5\text{,}\) \(A(5) = -2\text{,}\) \(A(6) = -0.5\text{.}\)

5. Use your work in (a)-(d) appropriately.

6. \(B(x) = A(x) + \frac{1}{2}\text{.}\)

Solution

1. \(A\) is accumulating positive signed area wherever \(g\) is positive, and thus \(A\) is increasing on \((0,1.5)\text{,}\) \((4,6)\text{;}\) \(A\) is accumulating negative signed area and therefore decreasing wherever \(g\) is negative, which occurs on \((1.5,4)\text{.}\)

2. Here we want to consider where \(A\) is changing at an increasing rate (concave up) or changing at a decreasing rate (concave down). On \((0,1)\) and \((4,5)\text{,}\) \(A\) is increasing, and we can also see that since \(g\) is increasing, \(A\) is increasing at an increasing rate. Similarly, on \((3,4)\) (where \(g\) is negative so \(A\) is decreasing), since \(g\) is increasing it follows that \(A\) is decreasing at an increasing rate. Thus, \(A\) is concave up on \((0,1)\) and \((3,5)\text{.}\) Analogous reasoning shows that \(A\) is concave down on \((1,3)\) and \((5,6)\text{.}\)

3. Based on our work in (a), we see that \(A\) changes from increasing to decreasing at \(x = 1.5\text{,}\) and thus \(A\) has a relative maximum there. Similarly, \(A\) has a relative minimum at \(x = 4\text{.}\)

4. Using the fact that \(g\) is piecewise linear and the definition of \(A\text{,}\) we find that \(A(0) = \int_1^0 g(t) \, dt = -\int_0^1 g(t) \, dt = -\frac{1}{2}\text{;}\) \(A(1) = \int_1^1 g(t) \, dt = 0\text{;}\) \(A(2) = \int_1^2 g(t) \, dt = 0\text{.}\) Analogous reasoning shows that \(A(3) = -2\text{,}\) \(A(4) = -3.5\text{,}\) \(A(5) = -2\text{,}\) \(A(6) = -0.5\text{.}\)

5. Use your work in (a)-(d) appropriately.

6. Note that \(B(0) = 0\text{,}\) while \(A(0) = -\frac{1}{2}\text{.}\) Likewise, \(B(1) = \frac{1}{2}\text{,}\) while \(A(1) = 0\text{.}\) Indeed, we can see that for any value of \(x\text{,}\) \(B(x) = A(x) + \frac{1}{2}\text{.}\)

Subsection 5.1.4 Summary

• Given the graph of a function \(f\text{,}\) we can construct the graph of its antiderivative \(F\) provided that (a) we know a starting value of \(F\text{,}\) say \(F(a)\text{,}\) and (b) we can evaluate the integral \(\int_a^b f(x) \, dx\) exactly for relevant choices of \(a\) and \(b\text{.}\) For instance, if we wish to know \(F(3)\text{,}\) we can compute \(F(3) = F(a) + \int_a^3 f(x) \, dx\text{.}\) When we combine this information about the function values of \(F\) together with our understanding of how the behavior of \(F' = f\) affects the overall shape of \(F\text{,}\) we can develop a completely accurate graph of the antiderivative \(F\text{.}\)

• Because the derivative of a constant is zero, if \(F\) is an antiderivative of \(f\text{,}\) it follows that \(G(x) = F(x) + C\) will also be an antiderivative of \(f\text{.}\) Moreover, any two antiderivatives of a function \(f\) differ precisely by a constant. Thus, any function with at least one antiderivative in fact has infinitely many, and the graphs of any two antiderivatives will differ only by a vertical translation.

• Given a function \(f\text{,}\) the rule \(A(x) = \int_a^x f(t) \, dt\) defines a new function \(A\) that measures the net-signed area bounded by \(f\) on the interval \([a,x]\text{.}\) We call the function \(A\) the integral function corresponding to \(f\text{.}\)