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Section 4.2 Riemann Sums

In Section 4.1, we learned that if an object moves with positive velocity \(v\text{,}\) the area between \(y = v(t)\) and the \(t\)-axis over a given time interval tells us the distance traveled by the object over that time period. If \(v(t)\) is sometimes negative and we view the area of any region below the \(t\)-axis as having an associated negative sign, then the sum of these signed areas tells us the moving object's change in position over a given time interval.

For instance, for the velocity function given in Figure 4.2.1, if the areas of shaded regions are \(A_1\text{,}\) \(A_2\text{,}\) and \(A_3\) as labeled, then the total distance \(D\) traveled by the moving object on \([a,b]\) is

\begin{equation*} D = A_1 + A_2 + A_3\text{,} \end{equation*}

while the total change in the object's position on \([a,b]\) is

\begin{equation*} s(b) - s(a) = A_1 - A_2 + A_3\text{.} \end{equation*}
Figure 4.2.1. A velocity function that is sometimes negative.

Because the motion is in the negative direction on the interval where \(v(t) \lt 0\text{,}\) we subtract \(A_2\) to determine the object's total change in position.

Of course, finding \(D\) and \(s(b)-s(a)\) for the graph in Figure 4.2.1 presumes that we can actually find the areas \(A_1\text{,}\) \(A_2\text{,}\) and \(A_3\text{.}\) So far, we have worked with velocity functions that were either constant or linear, so that the area bounded by the velocity function and the horizontal axis is a combination of rectangles and triangles, and we can find the area exactly. But when the curve bounds a region that is not a familiar geometric shape, we cannot find its area exactly. Indeed, this is one of our biggest goals in Chapter 4: to learn how to find the exact area bounded between a curve and the horizontal axis for as many different types of functions as possible.

In Activity 4.1.2, we approximated the area under a nonlinear velocity function using rectangles. In the following preview activity, we consider three different options for the heights of the rectangles we will use.

Preview Activity 4.2.1.

A person walking along a straight path has her velocity in miles per hour at time \(t\) given by the function \(v(t) = 0.25t^3-1.5t^2+3t+0.25\text{,}\) for times in the interval \(0 \le t \le 2\text{.}\) The graph of this function is also given in each of the three diagrams in Figure 4.2.2.

Figure 4.2.2. Three approaches to estimating the area under \(y = v(t)\) on the interval \([0,2]\text{.}\)

Note that in each diagram, we use four rectangles to estimate the area under \(y = v(t)\) on the interval \([0,2]\text{,}\) but the method by which the four rectangles' respective heights are decided varies among the three individual graphs.

  1. How are the heights of rectangles in the left-most diagram being chosen? Explain, and hence determine the value of

    \begin{equation*} S = A_1 + A_2 + A_3 + A_4 \end{equation*}

    by evaluating the function \(y = v(t)\) at appropriately chosen values and observing the width of each rectangle. Note, for example, that

    \begin{equation*} A_3 = v(1) \cdot \frac{1}{2} = 2 \cdot \frac{1}{2} = 1\text{.} \end{equation*}

  2. Explain how the heights of rectangles are being chosen in the middle diagram and find the value of

    \begin{equation*} T = B_1 + B_2 + B_3 + B_4\text{.} \end{equation*}

  3. Likewise, determine the pattern of how heights of rectangles are chosen in the right-most diagram and determine

    \begin{equation*} U = C_1 + C_2 + C_3 + C_4\text{.} \end{equation*}

  4. Of the estimates \(S\text{,}\) \(T\text{,}\) and \(U\text{,}\) which do you think is the best approximation of \(D\text{,}\) the total distance the person traveled on \([0,2]\text{?}\) Why?

Subsection 4.2.1 Sigma Notation

We have used sums of areas of rectangles to approximate the area under a curve. Intuitively, we expect that using a larger number of thinner rectangles will provide a better estimate for the area. Consequently, we anticipate dealing with sums of a large number of terms. To do so, we introduce sigma notation, named for the Greek letter \(\Sigma\text{,}\) which is the capital letter \(S\) in the Greek alphabet.

For example, say we are interested in the sum

\begin{equation*} 1 + 2 + 3 + \cdots + 100\text{,} \end{equation*}

the sum of the first 100 natural numbers. In sigma notation we write

\begin{equation*} \sum_{k=1}^{100} k = 1 + 2 + 3 + \cdots + 100\text{.} \end{equation*}

We read the symbol \(\sum_{k=1}^{100} k\) as “the sum from \(k\) equals 1 to 100 of \(k\text{.}\)” The variable \(k\) is called the index of summation, and any letter can be used for this variable. The pattern in the terms of the sum is denoted by a function of the index; for example,

\begin{equation*} \sum_{k=1}^{10} (k^2 + 2k) = (1^2 + 2\cdot 1) + (2^2 + 2\cdot 2) + (3^2 + 2\cdot 3) + \cdots + (10^2 + 2\cdot 10)\text{,} \end{equation*}

and more generally,

\begin{equation*} \sum_{k=1}^n f(k) = f(1) + f(2) + \cdots + f(n)\text{.} \end{equation*}

Sigma notation allows us to vary easily the function being used to describe the terms in the sum, and to adjust the number of terms in the sum simply by changing the value of \(n\text{.}\) We test our understanding of this new notation in the following activity.

Activity 4.2.2.

For each sum written in sigma notation, write the sum long-hand and evaluate the sum to find its value. For each sum written in expanded form, write the sum in sigma notation.

  1. \(\sum_{k=1}^{5} (k^2 + 2)\)

  2. \(\sum_{i=3}^{6} (2i-1)\)

  3. \(3 + 7 + 11 + 15 + \cdots + 27\)

  4. \(4 + 8 + 16 + 32 + \cdots + 256\)

  5. \(\sum_{i=1}^{6} \frac{1}{2^i}\)

Hint

  1. Observe that when \(k = 1\text{,}\) \(k^2 + 2 = 1^2 + 2 = 3\text{.}\) This is the first term in the sum.

  2. Note that this sum starts at \(i = 3\text{.}\)

  3. Since the terms in the sum increase by 4, try a function \(f(k)\) that somehow involves \(4k\text{.}\)

  4. What pattern do you observe in the terms of the sum?

  5. Write every term in the sum as a fraction with denominator \(2^6 = 64\text{.}\)

Answer

  1. \(64 \)

  2. \(32 \)

  3. \begin{equation*} 3 + 7 + 11 + 15 + \cdots + 27 = \sum_{k=1}^{7} 4k-1\text{.} \end{equation*}
  4. \begin{equation*} 4 + 8 + 16 + 32 + \cdots + 256 = \sum{i=2}^{8} 2^i\text{.} \end{equation*}
  5. \begin{equation*} \sum_{i=1}^{6} \frac{1}{2^i} = \frac{63}{64}\text{.} \end{equation*}
Solution
  1. \begin{align*} \sum_{k=1}^{5} (k^2 + 2) =\mathstrut \amp (1^2 + 2) + (2^2 + 2) + (3^2 + 2) + (4^2 + 2) + (5^2 + 2)\\ =\mathstrut \amp 3 + 6 + 11 + 17 + 27\\ =\mathstrut \amp 64 \end{align*}
  2. \begin{align*} \sum_{i=3}^{6} (2i-1) =\mathstrut \amp (2 \cdot 3 - 1) + (2 \cdot 4- 1) + (2 \cdot 5 - 1) + (2 \cdot 6 - 1)\\ =\mathstrut \amp 5 + 7 + 9 + 11\\ =\mathstrut \amp 32 \end{align*}

  3. Observe that each term in the sum

    \begin{equation*} 3 + 7 + 11 + 15 + \cdots + 27 \end{equation*}

    differs from the previous term by 4. If we view \(4\) as \(4 = 4 \cdot 1 - 1\) and \(7\) as \(7 = 4 \cdot 2 - 1\text{,}\) we see that the pattern may be represented through the function \(f(k) = 4k-1\text{,}\) so that

    \begin{equation*} 3 + 7 + 11 + 15 + \cdots + 27 = \sum_{k=1}^{7} 4k-1\text{.} \end{equation*}

    We note that \(k=7\) is the end value of the index since \(4 \cdot 7 = 28\text{.}\)

  4. The sum \(4 + 8 + 16 + 32 + \cdots + 256\) is a sum of powers of \(2\text{,}\) which we can express in sigma notation as

    \begin{equation*} 4 + 8 + 16 + 32 + \cdots + 256 = \sum{i=2}^{8} 2^i\text{.} \end{equation*}
  5. \begin{align*} \sum_{i=1}^{6} \frac{1}{2^i} =\mathstrut \amp \frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{2^6}\\ =\mathstrut \amp \frac{32}{64} + \frac{16}{64} + \frac{8}{64} + \frac{4}{64} + \frac{2}{64} + \frac{1}{64}\\ =\mathstrut \amp \frac{63}{64}\text{.} \end{align*}

Subsection 4.2.2 Riemann Sums

When a moving body has a positive velocity function \(y = v(t)\) on a given interval \([a,b]\text{,}\) the area under the curve over the interval gives the total distance the body travels on \([a,b]\text{.}\) We are also interested in finding the exact area bounded by \(y = f(x)\) on an interval \([a,b]\text{,}\) regardless of the meaning or context of the function \(f\text{.}\) For now, we continue to focus on finding an accurate estimate of this area by using a sum of the areas of rectangles. Unless otherwise indicated, we assume that \(f\) is continuous and non-negative on \([a,b]\text{.}\)

The first choice we make in such an approximation is the number of rectangles.

Figure 4.2.3. Subdividing the interval \([a,b]\) into \(n\) subintervals of equal length \(\Delta x\text{.}\)

If we desire \(n\) rectangles of equal width to subdivide the interval \([a,b]\text{,}\) then each rectangle must have width \(\Delta x = \frac{b-a}{n}\text{.}\) We let \(x_0 = a\text{,}\) \(x_n = b\text{,}\) and define \(x_{i} = a + i\Delta x\text{,}\) so that \(x_1 = x_0 + \Delta x\text{,}\) \(x_2 = x_0 + 2 \Delta x\text{,}\) and so on, as pictured in Figure 4.2.3.

We use each subinterval \([x_i, x_{i+1}]\) as the base of a rectangle, and next choose the height of the rectangle on that subinterval. There are three standard choices: we can use the left endpoint of each subinterval, the right endpoint of each subinterval, or the midpoint of each. These are precisely the options encountered in Preview Activity 4.2.1 and seen in Figure 4.2.2. We next explore how these choices can be described in sigma notation.

Consider an arbitrary positive function \(f\) on \([a,b]\) with the interval subdivided as shown in Figure 4.2.3, and choose to use left endpoints. Then on each interval \([x_{i}, x_{i+1}]\text{,}\) the area of the rectangle formed is given by

\begin{equation*} A_{i+1} = f(x_i) \cdot \Delta x\text{,} \end{equation*}

as seen in Figure 4.2.4.

Figure 4.2.4. Subdividing the interval \([a,b]\) into \(n\) subintervals of equal length \(\Delta x\) and approximating the area under \(y = f(x)\) over \([a,b]\) using left rectangles.

If we let \(L_n\) denote the sum of the areas of these rectangles, we see that

\begin{align*} L_n =\mathstrut \amp A_1 + A_2 + \cdots + A_{i+1} + \cdots + A_n\\ =\mathstrut \amp f(x_0) \cdot \Delta x + f(x_1) \cdot \Delta x + \cdots + f(x_i) \cdot \Delta x + \cdots + f(x_{n-1}) \cdot \Delta x\text{.} \end{align*}

In the more compact sigma notation, we have

\begin{equation*} L_n = \sum_{i = 0}^{n-1} f(x_i) \Delta x\text{.} \end{equation*}

Note that since the index of summation begins at \(0\) and ends at \(n-1\text{,}\) there are indeed \(n\) terms in this sum. We call \(L_n\) the left Riemann sum for the function \(f\) on the interval \([a,b]\text{.}\)

To see how the Riemann sums for right endpoints and midpoints are constructed, we consider Figure 4.2.5.

Figure 4.2.5. Riemann sums using right endpoints and midpoints.

For the sum with right endpoints, we see that the area of the rectangle on an arbitrary interval \([x_i, x_{i+1}]\) is given by \(B_{i+1} = f(x_{i+1}) \cdot \Delta x\text{,}\) and that the sum of all such areas of rectangles is given by

\begin{align*} R_n =\mathstrut \amp B_1 + B_2 + \cdots + B_{i+1} + \cdots + B_n\\ =\mathstrut \amp f(x_1) \cdot \Delta x + f(x_2) \cdot \Delta x + \cdots + f(x_{i+1}) \cdot \Delta x + \cdots + f(x_{n}) \cdot \Delta x\\ =\mathstrut \amp \sum_{i=1}^{n} f(x_i) \Delta x\text{.} \end{align*}

We call \(R_n\) the right Riemann sum for the function \(f\) on the interval \([a,b]\text{.}\)

For the sum that uses midpoints, we introduce the notation

\begin{equation*} \overline{x}_{i+1} = \frac{x_{i} + x_{i+1}}{2} \end{equation*}

so that \(\overline{x}_{i+1}\) is the midpoint of the interval \([x_i, x_{i+1}]\text{.}\) For instance, for the rectangle with area \(C_1\) in Figure 4.2.5, we now have

\begin{equation*} C_1 = f(\overline{x}_1) \cdot \Delta x\text{.} \end{equation*}

Hence, the sum of all the areas of rectangles that use midpoints is

\begin{align*} M_n =\mathstrut \amp C_1 + C_2 + \cdots + C_{i+1} + \cdots + C_n\\ =\mathstrut \amp f(\overline{x_1}) \cdot \Delta x + f(\overline{x_2}) \cdot \Delta x + \cdots + f(\overline{x}_{i+1}) \cdot \Delta x + \cdots + f(\overline{x}_{n}) \cdot \Delta x\\ =\mathstrut \amp \sum_{i=1}^{n} f(\overline{x}_i) \Delta x\text{,} \end{align*}

and we say that \(M_n\) is the middle Riemann sum for \(f\) on \([a,b]\text{.}\)

Thus, we have two variables to explore: the number of rectangles and the height of each rectangle. We can explore these choices dynamically, and the applet 1  found at http://gvsu.edu/s/a9 is a particularly useful one. There we see the image shown in Figure 4.2.6, but with the opportunity to adjust the slider bars for the heights and the number of rectangles.

Marc Renault, Geogebra Calculus Applets.
Figure 4.2.6. A snapshot of the applet found at http://gvsu.edu/s/a9.

By moving the sliders, we can see how the heights of the rectangles change as we consider left endpoints, midpoints, and right endpoints, as well as the impact that a larger number of narrower rectangles has on the approximation of the exact area bounded by the function and the horizontal axis.

When \(f(x) \ge 0\) on \([a,b]\text{,}\) each of the Riemann sums \(L_n\text{,}\) \(R_n\text{,}\) and \(M_n\) provides an estimate of the area under the curve \(y = f(x)\) over the interval \([a,b]\text{.}\) We also recall that in the context of a nonnegative velocity function \(y = v(t)\text{,}\) the corresponding Riemann sums approximate the distance traveled on \([a,b]\) by a moving object with velocity function \(v\text{.}\)

There is a more general way to think of Riemann sums, and that is to allow any choice of where the function is evaluated to determine the rectangle heights. Rather than saying we'll always choose left endpoints, or always choose midpoints, we simply say that a point \(x_{i+1}^*\) will be selected at random in the interval \([x_i, x_{i+1}]\) (so that \(x_i \le x_{i+1}^* \le x_{i+1}\)). The Riemann sum is then given by

\begin{equation*} f(x_1^*) \cdot \Delta x + f(x_2^*) \cdot \Delta x + \cdots + f(x_{i+1}^*) \cdot \Delta x + \cdots + f(x_n^*) \cdot \Delta x = \sum_{i=1}^{n} f(x_i^*) \Delta x\text{.} \end{equation*}

At http://gvsu.edu/s/a9, the applet noted earlier and referenced in Figure 4.2.6, by unchecking the “relative” box at the top left, and instead checking “random,” we can easily explore the effect of using random point locations in subintervals on a Riemann sum. In computational practice, we most often use \(L_n\text{,}\) \(R_n\text{,}\) or \(M_n\text{,}\) while the random Riemann sum is useful in theoretical discussions. In the following activity, we investigate several different Riemann sums for a particular velocity function.

Activity 4.2.3.

Suppose that an object moving along a straight line path has its velocity in feet per second at time \(t\) in seconds given by \(v(t) = \frac{2}{9}(t-3)^2 + 2\text{.}\)

  1. Carefully sketch the region whose exact area will tell you the value of the distance the object traveled on the time interval \(2 \le t \le 5\text{.}\)

  2. Estimate the distance traveled on \([2,5]\) by computing \(L_4\text{,}\) \(R_4\text{,}\) and \(M_4\text{.}\)

  3. Does averaging \(L_4\) and \(R_4\) result in the same value as \(M_4\text{?}\) If not, what do you think the average of \(L_4\) and \(R_4\) measures?

  4. For this question, think about an arbitrary function \(f\text{,}\) rather than the particular function \(v\) given above. If \(f\) is positive and increasing on \([a,b]\text{,}\) will \(L_n\) over-estimate or under-estimate the exact area under \(f\) on \([a,b]\text{?}\) Will \(R_n\) over- or under-estimate the exact area under \(f\) on \([a,b]\text{?}\) Explain.

Hint

  1. Note that \(y = v(t)\) is a parabola with vertex \((3,2)\text{.}\)

  2. Recall the formulas for \(L_n\text{,}\) \(R_n\text{,}\) and \(M_n\text{.}\)

  3. Think about what the average of \(L_1\) and \(R_1\) measures.

  4. Consider carefully the role of endpoints in generating \(L_n\) and \(R_n\text{.}\)

Answer

  2. \(L_4 = \frac{311}{48} \approx 6.47917\text{,}\) \(R_4 = \frac{335}{48} \approx 6.97917\text{,}\) and \(M_4 = \frac{637}{96} \approx 6.63542\text{.}\)

  3. \begin{equation*} \frac{L_4 + M_4}{2} = \frac{646}{96} \ne \frac{637}{96} = M_4\text{.} \end{equation*}

  4. \(L_n\) is an under-estimate; \(R_n\) is an over-estimates.

Solution

  1. The region whose exact area tells us the value of the distance the object traveled on the time interval \(2 \le t \le 5\) is shown below.

  2. \(L_4 = \frac{311}{48} \approx 6.47917\text{,}\) \(R_4 = \frac{335}{48} \approx 6.97917\text{,}\) and \(M_4 = \frac{637}{96} \approx 6.63542\text{.}\)

  3. The average of \(L_4\) and \(R_4\) is

    \begin{equation*} \frac{L_4 + M_4}{2} = \frac{311+335}{96} = \frac{646}{96} \ne \frac{637}{96} = M_4\text{.} \end{equation*}

    This average actually measures what would result from using four trapezoids, rather than rectangles, to estimate the area on each subinterval. One reason this is so is because the area of a trapezoid is the average of the bases times the width, and the “bases” are given by the function values at the left and right endpoints.

  4. If \(f\) is positive and increasing on \([a,b]\text{,}\) \(L_n\) will under-estimate the exact area under \(f\) on \([a,b]\text{.}\) Because \(f\) is increasing, its value at the left endpoint of any subinterval will be lower than every other function value in the interval, and thus the rectangle with that height lies exclusively below the curve. In a similar way, \(R_n\) over-estimates the exact area under \(f\) on \([a,b]\text{.}\)

Subsection 4.2.3 When the function is sometimes negative

For a Riemann sum such as

\begin{equation*} L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x\text{,} \end{equation*}

we can of course compute the sum even when \(f\) takes on negative values. We know that when \(f\) is positive on \([a,b]\text{,}\) a Riemann sum estimates the area bounded between \(f\) and the horizontal axis over the interval.

Figure 4.2.7. At left and center, two left Riemann sums for a function \(f\) that is sometimes negative; at right, the areas bounded by \(f\) on the interval \([a,d]\text{.}\)

For the function pictured in the first graph of Figure 4.2.7, a left Riemann sum with 12 subintervals over \([a,d]\) is shown. The function is negative on the interval \(b \le x \le c\text{,}\) so at the four left endpoints that fall in \([b,c]\text{,}\) the terms \(f(x_i) \Delta x\) are negative. This means that those four terms in the Riemann sum produce an estimate of the opposite of the area bounded by \(y = f(x)\) and the \(x\)-axis on \([b,c]\text{.}\)

In the middle graph of Figure 4.2.7, we see that by increasing the number of rectangles the approximation of the area (or the opposite of the area) bounded by the curve appears to improve.

In general, any Riemann sum of a continuous function \(f\) on an interval \([a,b]\) approximates the difference between the area that lies above the horizontal axis on \([a,b]\) and under \(f\) and the area that lies below the horizontal axis on \([a,b]\) and above \(f\text{.}\) In the notation of Figure 4.2.7, we may say that

\begin{equation*} L_{24} \approx A_1 - A_2 + A_3\text{,} \end{equation*}

where \(L_{24}\) is the left Riemann sum using 24 subintervals shown in the middle graph. \(A_1\) and \(A_3\) are the areas of the regions where \(f\) is positive, and \(A_2\) is the area where \(f\) is negative. We will call the quantity \(A_1 - A_2 + A_3\) the net signed area bounded by \(f\) over the interval \([a,d]\text{,}\) where by the phrase “signed area” we indicate that we are attaching a minus sign to the areas of regions that fall below the horizontal axis.

Finally, we recall that if the function \(f\) represents the velocity of a moving object, the sum of the areas bounded by the curve tells us the total distance traveled over the relevant time interval, while the net signed area bounded by the curve computes the object's change in position on the interval.

Activity 4.2.4.

Suppose that an object moving along a straight line path has its velocity \(v\) (in feet per second) at time \(t\) (in seconds) given by

\begin{equation*} v(t) = \frac{1}{2}t^2 - 3t + \frac{7}{2}\text{.} \end{equation*}

  1. Compute \(M_5\text{,}\) the middle Riemann sum, for \(v\) on the time interval \([1,5]\text{.}\) Be sure to clearly identify the value of \(\Delta t\) as well as the locations of \(t_0\text{,}\) \(t_1\text{,}\) \(\cdots\text{,}\) \(t_5\text{.}\) In addition, provide a careful sketch of the function and the corresponding rectangles that are being used in the sum.

  2. Building on your work in (a), estimate the total change in position of the object on the interval \([1,5]\text{.}\)

  3. Building on your work in (a) and (b), estimate the total distance traveled by the object on \([1,5]\text{.}\)

  4. Use appropriate computing technology 2  to compute \(M_{10}\) and \(M_{20}\text{.}\) What exact value do you think the middle sum eventually approaches as \(n\) increases without bound? What does that number represent in the physical context of the overall problem?

For instance, consider the applet at http://gvsu.edu/s/a9 and change the function and adjust the locations of the blue points that represent the interval endpoints \(a\) and \(b\text{.}\)
Hint

  1. Note that \(\Delta t = \frac{5-1}{5}\text{.}\)

  2. Change in position is tied to the net signed area bounded by the velocity function.

  3. Think about how total distance is different from change in position when the velocity is sometimes negative.

  4. Besides the noted applet, computer algebra systems such as Maple and Mathematica offer this utility.

Answer

  1. \(M_5 = -\frac{36}{25} = -1.44\)

  2. The change in position is approximately \(-1.44\) feet.

  3. \(D \approx 2.336\text{.}\)

  4. \(-\frac{4}{3}\) is the object's total change in position on \([1,5]\text{.}\)

Solution

  1. For this Riemann sum with five subintervals, \(\Delta t = \frac{5-1}{5} = \frac{4}{5}\text{,}\) so \(t_0 = 1\text{,}\) \(t_1 = 1.8\text{,}\) \(t_2 = 2.6\text{,}\) \(t_3 = 3.4\text{,}\) \(t_4 = 4.2\) and \(t_5 = 4\text{.}\) It follows that

    \begin{align*} M_5 =\mathstrut \amp v(1.4) \Delta t + v(2.2) \Delta t + v(3.0) \Delta t + v(3.8) \Delta t + v(4.6)\\ =\mathstrut \amp \frac{7}{25} \cdot \frac{4}{5} - \frac{17}{25} \cdot \frac{4}{5} - 1 \cdot \frac{4}{5} - \frac{17}{25} \cdot \frac{4}{5} + \frac{7}{25} \cdot \frac{4}{5}\\ =\mathstrut \amp -\frac{36}{25} = -1.44 \end{align*}

    A sketch is shown below.

  2. Since the net signed area bounded by \(v\) on \([1,5]\) represents the total change in position of the object on the interval \([1,5]\text{,}\) it follows that \(M_5\) estimates the total change in position. Hence, the change in position is approximately \(-1.44\) feet.

  3. To estimate the total distance traveled by the object on \([1,5]\text{,}\) we have to calculate the total area between the curve and the \(t\)-axis. Thus,

    \begin{equation*} D \approx \frac{7}{25} \cdot \frac{4}{5} + \frac{17}{25} \cdot \frac{4}{5} + 1 \cdot \frac{4}{5} + \frac{17}{25} \cdot \frac{4}{5} + \frac{7}{25} \cdot \frac{4}{5} = \frac{292}{125} \approx 2.336\text{.} \end{equation*}

  4. Using appropriate technology, \(M_{10} = -1.36\) and \(M_{20} = -1.34\text{.}\) Further calculations suggest that \(M_n \to -\frac{4}{3} = -1.\overline{33}\) as \(n \to \infty\text{,}\) and this number represents the object's total change in position on \([1,5]\text{.}\)

Subsection 4.2.4 Summary

  • A Riemann sum is simply a sum of products of the form \(f(x_i^*) \Delta x\) that estimates the area between a positive function and the horizontal axis over a given interval. If the function is sometimes negative on the interval, the Riemann sum estimates the difference between the areas that lie above the horizontal axis and those that lie below the axis.

  • The three most common types of Riemann sums are left, right, and middle sums, but we can also work with a more general Riemann sum. The only difference among these sums is the location of the point at which the function is evaluated to determine the height of the rectangle whose area is being computed. For a left Riemann sum, we evaluate the function at the left endpoint of each subinterval, while for right and middle sums, we use right endpoints and midpoints, respectively.

  • The left, right, and middle Riemann sums are denoted \(L_n\text{,}\) \(R_n\text{,}\) and \(M_n\text{,}\) with formulas

    \begin{align*} L_n = f(x_0) \Delta x + f(x_1) \Delta x + \cdots + f(x_{n-1}) \Delta x \amp= \sum_{i = 0}^{n-1} f(x_i) \Delta x,\\ R_n = f(x_1) \Delta x + f(x_2) \Delta x + \cdots + f(x_{n}) \Delta x \amp= \sum_{i = 1}^{n} f(x_i) \Delta x,\\ M_n = f(\overline{x}_1) \Delta x + f(\overline{x}_2) \Delta x + \cdots + f(\overline{x}_{n}) \Delta x \amp= \sum_{i = 1}^{n} f(\overline{x}_i) \Delta x\text{,} \end{align*}

    where \(x_0 = a\text{,}\) \(x_i = a + i\Delta x\text{,}\) and \(x_n = b\text{,}\) using \(\Delta x = \frac{b-a}{n}\text{.}\) For the midpoint sum, \(\overline{x}_{i} = (x_{i-1} + x_i)/2\text{.}\)

Exercises 4.2.5 Exercises

1. Evaluating Riemann sums for a quadratic function.
2. Estimating distance traveled with a Riemann sum from data.
3. Writing basic Riemann sums.
4.

Consider the function \(f(x) = 3x + 4\text{.}\)

  1. Compute \(M_4\) for \(y=f(x)\) on the interval \([2,5]\text{.}\) Be sure to clearly identify the value of \(\Delta x\text{,}\) as well as the locations of \(x_0, x_1, \ldots, x_4\text{.}\) Include a careful sketch of the function and the corresponding rectangles being used in the sum.

  2. Use a familiar geometric formula to determine the exact value of the area of the region bounded by \(y = f(x)\) and the \(x\)-axis on \([2,5]\text{.}\)

  3. Explain why the values you computed in (a) and (b) turn out to be the same. Will this be true if we use a number different than \(n = 4\) and compute \(M_n\text{?}\) Will \(L_4\) or \(R_4\) have the same value as the exact area of the region found in (b)?

  4. Describe the collection of functions \(g\) for which it will always be the case that \(M_n\text{,}\) regardless of the value of \(n\text{,}\) gives the exact net signed area bounded between the function \(g\) and the \(x\)-axis on the interval \([a,b]\text{.}\)

Answer

  1. \(M_4 = 43.5\text{.}\)

  2. \(A = \frac{87}{2}\text{.}\)

  3. The rectangles with heights that come from the midpoint have the same area as the trapezoids that are formed by the function values at the two endpoints of each subinterval.

    \(M_n\) will give the exact area for any value of \(n\text{.}\) Neither \(L_n\) nor \(R_n\) will be exact for any \(n\text{.}\)

  4. For any linear function \(g\) of the form \(g(x) = mx + b\) such that \(g(x) \ge 0\) on the interval of interest.

Solution

  1. Since we are using \(4\) subintervals on \([2,5]\text{,}\) it follows that \(\triangle x = \frac{5-2}{4} = 0.75\text{.}\) Thus, \(x_0 = 2\text{,}\) \(x_1 = 2.75\text{,}\) \(x_2 = 3.5\text{,}\) \(x_3 = 4.25\text{,}\) and \(x_4 = 5\text{.}\) The midpoints of the respective subintervals are thus \(\overline{x_1} = \frac{2+2.75}{2} = 2.375\text{,}\) \(\overline{x_2} = 3.125\text{,}\) \(\overline{x_3} = 3.875\text{,}\) and \(\overline{x_4} = 4.625\text{.}\) Thus,

    \begin{align*} M_4 &= f(\overline{x_1}) \cdot \triangle x + f(\overline{x_2}) \cdot \triangle x + f(\overline{x_3}) \cdot \triangle x + f(\overline{x_4}) \cdot \triangle x\\ &= \left(f(2.375) + f(3.125) + f(3.875) + f(4.625) \right) \cdot (0.75)\\ &= \left(11.125 + 13.375 + 15.625 + 17.875 \right) \cdot (0.75)\\ &= 58 \cdot 0.75\\ &= 43.5 \end{align*}

    An overall sketch of the key parts of \(M_4\) is seen in the figure below.

  2. Observe that \(f(2) = 10\) and \(f(5) = 19\text{.}\) The region bounded by the function, the \(x\)-axis, and the endpoints is a trapezoid. Recalling that the area of a trapezoid is the average of the bases times the height, we see that the bounded region has area

    \begin{equation*} A = \frac{1}{2}(10 + 19)(5-2) = \frac{87}{2}\text{.} \end{equation*}

  3. The reason that \(M_4\) gives the exact area between the line and the \(x\)-axis is the rectangles with heights that come from the midpoint have the same area as the trapezoids that are formed by the function values at the two endpoints of each subinterval. We can also see this by observing that the ``extra area'' bounded by the midpoint rectangle exactly matches the ``missing area'' that is absent from the midpoint rectangle, as seen in the following figure. Since these two quantities balance out, the areas of the midpoint rectangles give us the exact area under the curve.

    Indeed, the value of \(M_n\) will give the exact area under \(g\) on \([2,5]\) for any value of \(n\text{.}\) But neither \(L_n\) nor \(R_n\) will be exact: \(L_n\) will underestimate the true area since every rectangle will lie completely under the function \(g\text{,}\) whereas \(R_n\) will overestimate because every rectangle lies above the function.

  4. For any linear function \(g\) of the form \(g(x) = mx + b\) such that \(g(x) \ge 0\) on the interval of interest, the midpoint rule will give the exact value of the area bounded by the function and the \(x\)-axis. This is true because the area of the trapezoid formed by the two endpoints on a given subinterval will be the same as the area of the rectangle whose height is given by the function at the midpoint of the subinterval.

5.

Let \(S\) be the sum given by

\begin{equation*} S = ((1.4)^2 + 1) \cdot 0.4 + ((1.8)^2 + 1) \cdot 0.4 + ((2.2)^2 + 1) \cdot 0.4 + ((2.6)^2 + 1) \cdot 0.4 +((3.0)^2 + 1) \cdot 0.4\text{.} \end{equation*}

  1. Assume that \(S\) is a right Riemann sum. For what function \(f\) and what interval \([a,b]\) is \(S\) this function's Riemann sum? Why?

  2. How does your answer to (a) change if \(S\) is a left Riemann sum? a middle Riemann sum?

  3. Suppose that \(S\) really is a right Riemann sum. What is geometric quantity does \(S\) approximate?

  4. Use sigma notation to write a new sum \(R\) that is the right Riemann sum for the same function, but that uses twice as many subintervals as \(S\text{.}\)

Answer

  1. \(f(x) = x^2 + 1\) on the interval \([1,3]\text{.}\)

  2. If \(S\) is a left Riemann sum, \(f(x) = x^2 + 1\) on the interval \([1.4, 3.4]\text{.}\) If \(S\) is a middle Riemann sum, \(f(x) = x^2 + 1\) on the interval \([1.2, 3.2]\text{.}\)

  3. The area under \(f(x) = x^2 + 1\) on \([1,3]\text{.}\)

  4. \(R_{10} = \sum_{i=1}^{10} \left( (1+0.2i)^2 + 1 \right) \cdot 0.2\text{.}\)

Solution

  1. The function is being evaluated at \(x_1 = 1.4\text{,}\) \(x_2 = 1.8\text{,}\) \(x_3 = 2.2\text{,}\) \(x_4 = 2.6\text{,}\) and \(x_5 = 3.0\text{.}\) Thus, since we know this is a right Riemann sum, we are using \(5\) subintervals with \(\triangle x = 0.4\text{,}\) and \(x_0 = 1.0\text{.}\) From the pattern, we observe that the function must be \(f(x) = x^2 + 1\) on the interval \([1,3]\text{.}\)

  2. If \(S\) is a left Riemann sum instead, then the \(x_i\)'s shift left: \(x_0 = 1.4\text{,}\) \(x_1 = 1.8\text{,}\) \(\ldots\text{,}\) \(x_5 = 3.4\text{.}\) In this setting, the function is identical, but we are considering it on the interval \([1.4, 3.4]\text{.}\)

    If \(S\) is a middle Riemann sum, then (for example) \(1.4\) is the midpoint of an interval \(\triangle x = 0.4\) units long, so \(x_0 = 1.2\) and \(x_1 = 1.6\text{.}\) Continuing that trend, we'd be considering the function \(f(x) = x^2 + 1\) on the interval \([1.2, 3.2]\text{.}\)

  3. \(S\) is an approximation of the area that lies under \(f(x) = x^2 + 1\) on \([1,3]\text{.}\)

  4. To use twice as many subintervals, we move to \(\triangle x = 0.2\) and thus \(x_1 = 1.2\text{,}\) \(x_2 = 1.4\text{,}\) and so on, with the general right endpoint being given by \(x_i = 1 + 0.2 \cdot i\text{.}\) We'll also have \(n = 10\text{,}\) so

    \begin{equation*} R_{10} = \sum_{i=1}^{10} \left( (1+0.2i)^2 + 1 \right) \cdot 0.2\text{.} \end{equation*}
6.

A car traveling along a straight road is braking and its velocity is measured at several different points in time, as given in the following table.

seconds, \(t\) \(0\) \(0.3\) \(0.6\) \(0.9\) \(1.2\) \(1.5\) \(1.8\)
Velocity in ft/sec, \(v(t)\) \(100\) \(88\) \(74\) \(59\) \(40\) \(19\) \(0\)
Table 4.2.8. Data for the braking car.

  1. Plot the given data on a set of axes with time on the horizontal axis and the velocity on the vertical axis.

  2. Estimate the total distance traveled during the car the time brakes using a middle Riemann sum with 3 subintervals.

  3. Estimate the total distance traveled on \([0,1.8]\) by computing \(L_6\text{,}\) \(R_6\text{,}\) and \(\frac{1}{2}(L_6 + R_6)\text{.}\)

  4. Assuming that \(v(t)\) is always decreasing on \([0,1.8]\text{,}\) what is the maximum possible distance the car traveled before it stopped? Why?

Answer

  2. \(M_3 = 99.6\) feet.

  3. \(L_6 = 114 \text{,}\)

    \begin{equation*} R_6 = 84\text{,} \end{equation*}

    and \(\frac{1}{2}(L_6 + R_6) = 99\text{.}\)

  4. \(114\) feet.

Solution

  1. A plot of the given data is shown in the figure below.

  2. For \(M_3\) with the given data, the midpoints are \(\overline{x_1} = 0.3\text{,}\) \(\overline{x_2} = 0.9\text{,}\) and \(\overline{x_3} = 1.5\text{,}\) and \(\triangle t = 0.6\text{,}\) so

    \begin{equation*} M_3 = f(0.3) \cdot 0.6 + f(0.9) \cdot 0.6 + f(1.5) \cdot 0.6 = 166 \cdot 0.6 = 99.6\text{.} \end{equation*}

    Therefore, the car travels approximately \(99.6\) feet over the \(1.8\) second interval it brakes.

  3. Using the standard formulas,

    \begin{equation*} L_6 = f(0) \cdot 0.3 + f(0.3) \cdot 0.3 + \cdots + f(1.5) \cdot 0.3 = 380 \cdot 0.3 = 114 \end{equation*}

    and

    \begin{equation*} R_6 = f(0.3) \cdot 0.3 + f(0.6) \cdot 0.3 + \cdots + f(1.8) \cdot 0.3 = 280 \cdot 0.3 = 84\text{.} \end{equation*}

    Thus,

    \begin{equation*} \frac{1}{2}(L_6 + R_6) = \frac{114+84}{2} = 99\text{.} \end{equation*}

    Each of these three quantities is an estimate of the total distance traveled: \(L_6\) is an overestimate and \(R_6\) is an underestimate.

  4. The maximum possible distance the car traveled before it stopped is \(114\) feet. This is because \(L_n\) is always an overestimate when the function is decreasinging on the interval.

7.

The rate at which pollution escapes a scrubbing process at a manufacturing plant increases over time as filters and other technologies become less effective. For this particular example, assume that the rate of pollution (in tons per week) is given by the function \(r\) that is pictured in Figure 4.2.9.

  1. Use the graph to estimate the value of \(M_4\) on the interval \([0,4]\text{.}\)

  2. What is the meaning of \(M_4\) in terms of the pollution discharged by the plant?

  3. Suppose that \(r(t) = 0.5 e^{0.5t}\text{.}\) Use this formula for \(r\) to compute \(L_5\) on \([0,4]\text{.}\)

  4. Determine an upper bound on the total amount of pollution that can escape the plant during the pictured four week time period that is accurate within an error of at most one ton of pollution.

Figure 4.2.9. The rate, \(r(t)\text{,}\) of pollution in tons per week.
Answer

  1. \(M_4 \approx 6.4\text{.}\)

  2. The total tonnage of pollution escaping the scrubbing process in the time interval \([0,4]\) weeks.

  3. \(L_5 \approx 5.19620599\text{.}\)

  4. \(6.4\) tons.

Solution

  1. We estimate the value of \(r\) at the midpoint of each subinterval and then multiply these values by the width (\(1\)) of each subinterval to estimate the areas of the rectangles shown in the following figure. We use the following approximations: \(r(0.5) \approx 0.6\text{,}\) \(r(1.5) \approx 1.1\text{,}\) \(r(2.5) \approx 1.8\text{,}\) \(r(3.5) \approx 2.9\text{,}\) so

    \begin{equation*} M_4 \approx (0.6)(1) + (1.1)(1) + (1.8)(1) + (2.9)(1) = 6.4\text{.} \end{equation*}

    \[\]

  2. Since \(r(t)\) is measured in tons/week and \(t\) in weeks, the units of \(r(t) \triangle t\) are tons. So \(M_4\) approximates the total tonnage of pollution escaping the scrubbing process in the time interval \([0,4]\) weeks.

  3. Now we are using \(5\) subintervals, so \(\triangle t = \frac{4}{5} = 0.8\text{.}\) We evaluate \(r\) at the left endpoint of each subinterval and multiply by \(\triangle t\)to determine the areas of the rectangles shown in the following figure. So

    \begin{align*} L_5 &= \left(0.5e^{(0.5)(0)}\right)(0.8) + \left(0.5e^{(0.5)(0.8)}\right)(0.8) + \left(0.5e^{(0.5)(1.6)}\right)(0.8)\\ &+ \left(0.5e^{(0.5)(2.4)}\right)(0.8) + \left(0.5e^{(0.5)(3.2)}\right)(0.8)\\ &\approx 5.19620599\text{.} \end{align*}

  4. Given that \(r\) is an increasing function, a right sum will always give us an overestimate of the total amount of pollution (in tons) on the interval \([0,4]\text{.}\) We use computational technology to compute a few right sums: \(R_{10} \approx 7.049244376\text{,}\) \(R_{50} \approx 6.517689073\text{,}\) \(R_{100} \approx 6.453159616\text{,}\) \(R_{500} \approx 6.401842731\text{.}\) These sums don't seem to change in the tens place as we make \(n\) sufficiently large, so an upper bound on the total amount of pollution on the interval \([0,4]\) that is accurate to within one ton is \(6.4\) tons.

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