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Section5.1Constructing Accurate Graphs of Antiderivatives

Motivating Questions
  • Given the graph of a function's derivative, how can we construct a completely accurate graph of the original function?

  • How many antiderivatives does a given function have? What do those antiderivatives all have in common?

  • Given a function \(f\text{,}\) how does the rule \(A(x) = \int_0^x f(t) \, dt\) define a new function \(A\text{?}\)

A recurring theme in our discussion of differential calculus has been the question: given information about the derivative of an unknown function \(f\text{,}\) how much information can we obtain about \(f\) itself? In Example2.109, the graph of \(y = f'(x)\) was known (along with the value of \(f\) at a single point) and we endeavored to sketch a possible graph of \(f\) near the known point. In Example3.5, we investigated how the first derivative test enables us to use information about \(f'\) to determine where the original function \(f\) is increasing and decreasing, as well as where \(f\) has relative extreme values. If we know a formula or graph of \(f'\text{,}\) we can find where the original function \(f\) is concave up and concave down by computing \(f''\text{.}\) Thus, knowing \(f'\) and \(f''\) enables us to understand the shape of the graph of \(f\text{.}\)

We returned to this question in even more detail in Section4.1. In that setting, we knew the instantaneous velocity of a moving object and worked to determine as much as possible about the object's position function. We found connections between the net signed area under the velocity function and the corresponding change in position of the function, and the Total Change Theorem further illuminated these connections between \(f'\) and \(f\text{,}\) showing that the total change in the value of \(f\) over an interval \([a,b]\) is determined by the net signed area bounded by \(f'\) and the \(x\)-axis on the same interval.

In what follows, we explore the situation where we possess an accurate graph of the derivative function along with a single value of the function \(f\text{.}\) From that information, we'd like to determine a graph of \(f\) that shows where \(f\) is increasing, decreasing, concave up, and concave down, and also provides an accurate function value at any point.

Example5.1

Suppose that the following information is known about a function \(f\text{:}\) the graph of its derivative, \(y = f'(x)\text{,}\) is given in Figure5.2. Further, assume that \(f'\) is piecewise linear (as pictured) and that for \(x \le 0\) and \(x \ge 6\text{,}\) \(f'(x) = 0\text{.}\) Finally, it is given that \(f(0) = 1\text{.}\)

Figure5.2At left, the graph of \(y = f'(x)\text{;}\) at right, axes for plotting \(y = f(x)\text{.}\)
  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?

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

  4. Recall that the Total Change Theorem tells us that

    \begin{equation*} f(1) - f(0) = \int_0^1 f'(x) \, dx\text{.} \end{equation*}

    What is the exact value of \(f(1)\text{?}\)

  5. Use the given information and similar reasoning to that in (d) to determine the exact value of \(f(2)\text{,}\) \(f(3)\text{,}\) \(f(4)\text{,}\) \(f(5)\text{,}\) and \(f(6)\text{.}\)

  6. 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 \(x \gt 6\text{.}\)

Hint
  1. Remember what values of the derivative result in the function being increasing and decreasing. Remember that the graph you are given is of the derivative, so positive values of the derivative are really positive \(y\) values (above the \(x\)-axis).

  2. Remember that the second derivative tells us about concavity. The second derivative of the function would be the derivative (slopes) of the function in the graph you were given.

  3. Remember that we could have relative minima or maxima where the derivative is zero. Determine where you have minima or maxima by looking at where you decided the function was increasing or decreasing in part (a).

  4. Remember that the integral represents the area under the curve.

  5. Same as above.

  6. Plot the points you found and use the remaining parts to fill in the details.

Answer
  1. Increasing: \((0,1.5)\cup(4,6)\)

    Decreasing: \((1.5,4)\)

  2. Concave up: \((0,1)\cup(3,5)\)

    Concave down: \((1,3)\cup (5,6)\)

  3. Relative minima: \(x=1.5, 6\)

    Relative maximum: \(x=4\)

  4. \(f(1)=1.5\)

  5. \(f(2)=1.5, f(3)=-0.5, f(4)=-2, f(5)=-0.5, f(6)=1\)

Solution
  1. To find where the function is increasing, we want to look at where the derivative function is positive (i.e., where the derivative is above the \(x\)-axis). Therefore, \(f(x)\) is increasing on \((0,1.5)\cup(4,6)\text{.}\)

    To find where the function is decreasing, we want to look at where the derivative function is negative (i.e., where the derivative is below the \(x\)-axis). Therefore, \(f(x)\) is decreasing on \((1.5,4)\text{.}\)

  2. To find where the function is concave up, we want to look at where the slope of the derivative is positive. Therefore, \(f(x)\) is concave up on \((0,1)\cup(3,5)\text{.}\)

    To find where the function is concave down, we want to look at where the slope of the derivative is negative. Therefore, \(f(x)\) is concave down on \((1,3)\cup (5,6)\text{.}\)

  3. To find where the function has relative minima, we want to look at where the derivative is zero. In order for the values to be a minimum, we need them to occur after the function increases and then it changes to decreasing. Therefore, there are relative minima at \(x=1.5, 6\text{.}\)

    Similarly, to find relative maxima, we look where the derivative is zero, and after the function decreases and then changes to increasing. Therefore, there is a relative maximum at \(x=4\text{.}\)

  4. The integral on the right side of the equation \(\int_0^1 f'(x) \, dx\) is the value of the area under the curve between 0 and 1. That area is given by \(1\cdot 1\cdot 0.5\text{.}\) Therefore, \(\int_0^1 f'(x) \, dx=0.5\text{.}\) Since \(f(0)=1\text{,}\) we have \(f(1)-1=0.5\text{.}\) Therefore, \(f(1)=1.5\)

  5. We follow a similar process to find the rest of the values: \(f(2)=1.5, f(3)=-0.5, f(4)=-2, f(5)=-0.5, f(6)=1.\)

    Note that in this case, it can be more accurate to solve for \(f(a)\) by solving the following equation in each case

    \begin{equation*} f(a)-f(0)=\int_0^a f'(x) \, dx\text{,} \end{equation*}

    since we are given the value of \(f(0)\text{.}\)

SubsectionConstructing the graph of an antiderivative

Example5.1 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

\begin{equation} F(b) = F(a) + \int_a^b f(x) \, dx\text{.}\label{E-FTCTCT}\tag{5.1} \end{equation}

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 \(\int_a^b f(x) \, dx\text{.}\)

Example5.3

Suppose that the function \(y = f(x)\) is given by the graph shown in Figure5.4, 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 \(x \ge 7\text{,}\) \(f(x) = 0\text{.}\)

Figure5.4At left, the graph of \(y = f(x)\text{.}\)
  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: \(G(x) = F(x) + 1\text{.}\)

SubsectionMultiple antiderivatives of a single function

In the final question of Example5.3, 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 Figure5.5, and suppose further that \(F\) is an antiderivative of \(f\) that satisfies \(F(0) = 1\text{.}\)

Figure5.5At left, the graph of \(y = f(x)\text{.}\) At right, three different antiderivatives of \(f\text{.}\)

Then, using Equation(5.1), we can compute

\begin{align*} F(1) &= F(0) + \int_0^1 f(x) \, dx\\ &= 1 + 0.5\\ &= 1.5\text{.} \end{align*}

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 Figure5.5.

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 Figure5.5. 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 Figure5.5.

This example demonstrates an important fact that holds more generally:

If \(G\) and \(H\) are both antiderivatives of a function \(f\text{,}\) then the function \(G - H\) must be constant.

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,

\begin{equation*} \frac{d}{dx}[ G(x) - H(x) ] = G'(x) - H'(x) = f(x) - f(x) = 0\text{.} \end{equation*}

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

\begin{equation*} 3 = \frac{1}{3}(2)^3 + C\text{,} \end{equation*}

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

Example5.6

For each of the following functions, sketch an accurate graph of the antiderivative that satisfies the given initial condition. In addition, sketch the graph of two additional antiderivatives of the given function, and state the corresponding initial conditions that each of them satisfy. If possible, find an algebraic formula for the antiderivative that satisfies the initial condition.

  1. Original function: \(g(x) = \left| x \right| - 1\text{;}\) initial condition: \(G(-1) = 0\text{;}\) interval for sketch: \([-2,2]\)

  2. Original function: \(h(x) = \sin(x)\text{;}\) initial condition: \(H(0) = 1\text{;}\) interval for sketch: \([0,4\pi]\)

  3. Original function: \(p(x) = \begin{cases}x^2, \amp \text{ if } 0 \lt x \le 1 \\ -(x-2)^2, \amp \text{ if } 1 \lt x \lt 2 \\ 0 \amp \text{ otherwise } \end{cases}\text{;}\) initial condition: \(P(0) = 1\text{;}\) interval for sketch: \([-1,3]\)

Hint
  1. Consider breaking \(g(x)\) into two cases: \(x\lt 0\) and \(x\ge 0\text{.}\) This will help you find a possible antiderivative (think about what function you can take the derivative of to get the original function or use areas under the curve). Then you can use the initial condition to help you find \(G(x)\text{.}\) Try varying the initial conditions to find other possible antiderivatives.

  2. Think about what function has \(sin(x)\) as its derivative. Then use the initial condition to solve for any possible shift.

  3. Consider each piece individually. Think about what function you can take the derivative of to get the original function. Then use the initial condition.

Answer
  1. \(H(x) = -\cos(x) + 2\text{.}\)

Solution
  1. A possible antiderivative \(G\) that satifies \(G(0) = 0\) is shown in the figure below, with a possible formula being \(G(x) = -\frac{1}{2}x^2 - x\) for \(x \lt 0\) and \(G(x) = \frac{1}{2}x^2 - x\) for \(x \ge 0\text{.}\) Other antiderivatives that satisfy \(F(0) = 2\) and \(H(0) = -1\) are also shown.

  2. Thinking both graphically and algebraically reveals that the antiderivative we seek is \(H(x) = -\cos(x) + 2\text{.}\) Any vertical shift of this function will have the same derivative, but will have a different \(y\)-intercept.

  3. A possible antiderivative \(P\) that satifies \(P(0) = 1\) is shown in the figure below.

SubsectionFunctions defined by integrals

Equation(5.1) 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,

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

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.

Definition5.7

If \(f\) is a continuous function, we define the corresponding integral function \(A\) according to the rule

\begin{equation} A(x) = \int_a^x f(t) \, dt\text{.}\label{E-intfxn}\tag{5.2} \end{equation}

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 Figure5.8, 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.

Figure5.8At left, the graph of the given function \(f\text{.}\) At right, the area function \(A(x) = \int_0^x f(t) \ dt\text{.}\)

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 applet1David Austin, Grand Valley State University that brings the static picture in Figure5.8 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 example that follows, we explore how the value of \(a\) affects the graph of the integral function.

Example5.9

Suppose that \(g\) is given by the graph at left in Figure5.10 and that \(A\) is the corresponding integral function defined by \(A(x) = \int_1^x g(t) \, dt\text{.}\)

Figure5.10At left, the graph of \(y = g(t)\text{;}\) at right, axes for plotting \(y = A(x)\text{,}\) where \(A\) is defined by the formula \(A(x) = \int_1^x g(t) \ dt\text{.}\)
  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 \(x\text{,}\) \(B(x) = A(x) + \frac{1}{2}\text{.}\)

SubsectionSummary

  • 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 \(f\text{.}\)

SubsectionSupplemental Videos

SubsectionExercises

A moving particle has its velocity given by the quadratic function \(v\) pictured in Figure5.11. In addition, it is given that \(A_1 = \frac{7}{6}\) and \(A_2 = -\frac{8}{3}\text{,}\) as well as that for the corresponding position function \(s\text{,}\) \(s(0) = 0.5\)

  1. Use the given information to determine \(s(1)\text{,}\) \(s(3)\text{,}\) \(s(5)\text{,}\) and \(s(6)\text{.}\)

  2. On what interval(s) is \(s\) increasing? On what interval(s) is \(s\) decreasing?

  3. On what interval(s) is \(s\) concave up? On what interval(s) is \(s\) concave down?

  4. Sketch an accurate, labeled graph of \(s\) on the axes at right in Figure5.11.

  5. Note that \(v(t) = -2 + \frac{1}{2}(t-3)^2\text{.}\) Find a formula for \(s\text{.}\)

Figure5.11At left, the given graph of \(v\text{.}\) At right, axes for plotting \(s\text{.}\)

A person exercising on a treadmill experiences different levels of resistance and thus burns calories at different rates, depending on the treadmill's setting. In a particular workout, the rate at which a person is burning calories is given by the piecewise constant function \(c\) pictured in Figure5.12. Note that the units on \(c\) are calories per minute.

Figure5.12At left, the given graph of \(c\text{.}\) At right, axes for plotting \(C\text{.}\)
  1. Let \(C\) be an antiderivative of \(c\text{.}\) What does the function \(C\) measure? What are its units?

  2. Assume that \(C(0) = 0\text{.}\) Determine the exact value of \(C(t)\) at the values \(t = 5, 10, 15, 20, 25, 30\text{.}\)

  3. Sketch an accurate graph of \(C\) on the axes provided at right in Figure5.12. Be certain to label the scale on the vertical axis.

  4. Determine a formula for \(C\) that does not involve an integral and is valid for \(5 \le t \le 10\text{.}\)

Consider the piecewise linear function \(f\) given in Figure5.13. Let the functions \(A\text{,}\) \(B\text{,}\) and \(C\) be defined by the rules \(A(x) = \int_{-1}^{x} f(t) \, dt\text{,}\) \(B(x) = \int_{0}^{x} f(t) \, dt\text{,}\) and \(C(x) = \int_{1}^{x} f(t) \, dt\text{.}\)

  1. For the values \(x = -1, 0, 1, \ldots, 6\text{,}\) make a table that lists corresponding values of \(A(x)\text{,}\) \(B(x)\text{,}\) and \(C(x)\text{.}\)

  2. On the axes provided in Figure5.13, sketch the graphs of \(A\text{,}\) \(B\text{,}\) and \(C\text{.}\)

  3. How are the graphs of \(A\text{,}\) \(B\text{,}\) and \(C\) related?

  4. How would you best describe the relationship between the function \(A\) and the function \(f\text{?}\)

Figure5.13At left, the given graph of \(f\text{.}\) At right, axes for plotting \(A\text{,}\) \(B\text{,}\) and \(C\text{.}\)