Recall the two-step procedure for finding the slope of the tangent line (the derivative): First approximate the slope using secant lines and then compute the limit of these approximations. In integral calculus, there are also two steps:

• First, approximate the area under the graph using rectangles, and then
• Compute the exact area (the integral) as the limit of these approximations.

To summarize,

EXAMPLE 1

Calculate \(R_4\) and \(R_6\) for \(f(x) = x^2\) on the interval \([1, 3]\).

Solution

Step 1. Determine \(\Delta x\) and the right endpoints.

To calculate \(R_4\), divide \([1, 3]\) into four subintervals of width \(\Delta x = \tfrac{3-1}{4}=\tfrac{1}{2}\).The right endpoints are the numbers \(a + j\Delta x = 1 + \tfrac{j}{2}\) for \(j = 1, 2, 3, 4\). They are spaced at intervals of \(\tfrac12\) beginning at \(\tfrac32\), so, as we see in Figure 5.5 (move the slider to display \(R_4\)), the right endpoints are \(\tfrac32,\tfrac42,\tfrac52,\tfrac62\).

Figure 5.5: A dynamical figure to investigate approximating the area under the given graph by left-endpoint, mid-point or right-endpoint approximations. Experiment!

Step 2. Calculate \(\Delta x\) times the sum of function values.

\(R_4\) is \(\Delta x\) times the sum of the function values at the right endpoints:

\(R_6\) is similar: \(\Delta x = \tfrac{3-1}{6} = \tfrac13\), and the right endpoints are spaced at intervals of \(\tfrac13\) beginning at \(\tfrac43\) and ending at \(3\), as in Figure 5.5 (move the slider to display \(R_6\)). Thus,

Linearity of Summations

• \(\sum\limits_{j=m}^n (a_j + b_j) =\sum\limits_{j=m}^n a_j + \sum\limits_{j=m}^n b_j \)
• \(\sum\limits_{j=m}^n C a_j = C \sum\limits_{j=m}^n a_j \) (\(C\) any constant)
• \(\sum\limits_{j=1}^n k = nk \) (\(k\) any constant and \(n\geq 1\))

REMINDER

EXAMPLE 2

Calculate \(R_6\), \(L_6\), and \(M_6\) for \(f(x)= x^{-1}\) on \([2, 4]\).

Solution In this case, \(\Delta x = \tfrac{b-a}N = \tfrac{4-2}6 = \tfrac13\). The general term in the summation for \(R_6\) and \(L_6\) is

Therefore (Figure 5.7),

Figure 5.7: \(L_6\) and \(R_6\) for \(f(x)= x^{-1}\) on \([2, 4]\). Note: the left-endpoint rectangles contain the right-endpoint rectangles since \(f(x)\) is decreasing and therefore \(L_6 > R_6\).

In \(L_6\), the sum begins at \(j=0\) and ends at \(j=5\):

The general term in \(M_6\) is

Summing up from \(j=1\) to \(j=6\), we obtain (Figure 5.8)

Figure 5.8: \(M_6\) for \(f(x)= x^{-1}\) on \([2, 4]\).

GRAPHICAL INSIGHT Monotonic Functions

Observe in Figure 5.7 that the left-endpoint rectangles for \(f(x)= x^{-1}\) extend above the graph and the right-endpoint rectangles lie below it. The exact area \(A\) must lie between \(R_6\) and \(L_6\), and so, according to the previous example, \(0.65\leq A\leq 0.74\). More generally, when \(f(x)\) is monotonic (increasing or decreasing), the exact area lies between \(R_N\) and \(L_N\) (Figure 5.9):

Figure 5.9: When \(f(x)\) is increasing, the left-endpoint rectangles lie below the graph and right-endpoint rectangles lie above it.

• \(f(x)\text{ increasing}\implies L_N\leq\text{area under graph}\leq R_N\)
• \(f(x)\text{ decreasing}\implies L_N\geq\text{area under graph}\geq R_N\)

In Theorem 1, it is not assumed that \(f(x)\leq 0\). If \(f(x)\) takes on negative values, the limit \(L\) no longer represents area under the graph, but we can interpret it as a sum and difference of areas, which is discussed in the next section.

THEOREM 1

If \(f(x)\) is continuous on \([a,b]\), then the endpoint and midpoint approximations approach one and the same limit as \(N\rightarrow\infty\). In other words, there is a value \(L\) such that

If \(f(x)\geq 0\), we define the area under the graph over \([a,b]\) to be \(L\).

CONCEPTUAL INSIGHT

In calculus, limits are used to define basic quantities that otherwise would not have a precise meaning. Theorem 1 allows us to define area as a limit \(L\) in much the same way that we define the slope of a tangent line as the limit of slopes of secant lines.

A method for proving power sum formulas is developed in Exercises 40–43 of Section 1.3. Formulas (3)–(5) can also be verified using the method of induction.

Power Sums

EXAMPLE 3

Find the area \(A\) under the graph of \(f(x)=x\) over \([0, 4]\) in three ways:

• (a) Using geometry
• (b) \(\lim\limits_{ N\rightarrow \infty} R_N\)
• (c) \(\lim\limits_{ N\rightarrow \infty} L_N\)

Solution The region under the graph is a right triangle with base \(b=4\) and height \(h = 4\) (Figure 5.11).

Figure 5.11: The right-endpoint approximations approach the area of the triangle.

• (a) By geometry, \(A= \tfrac12bh=(\tfrac12)(4)(4)=8\).
• (b) We compute this area again as a limit. Since \(\Delta x = \tfrac{b-a}N = \tfrac4N\)and \(f(x) = x\), \[ \begin{aligned} f(a+j\Delta x) & = f\left(0 + j\left(\frac4N\right)\right) = \frac{4j}N\\ R_N & =\Delta \sum\limits_{j=1}^N f(a+j\Delta x) = \frac4N \sum\limits_{j=1}^N \frac{4j}N = \frac{16}{N^2} \sum\limits_{j=1}^N j \end{aligned} \]

  REMINDER \[ \begin{aligned} R_N & = \Delta \sum\limits_{j=1}^N f(a+j\Delta x) \\ L_N & = \Delta \sum\limits_{j=0}^{N-1} f(a+j\Delta x) \\ \Delta c & = \frac{b-a}N \end{aligned} \]

  In the last equality, we factored out \(\tfrac4N\) from the sum. This is valid because \(\tfrac4N\) is a constant that does not depend on \(j\). Now use formula (3): \[ R_N = \frac{16}{N^2} \sum\limits_{j=1}^N j = \frac{16}{N^2} \underbrace{\left(\frac{N(N+1)}2\right)}_{\text{Formula for power sum}} = \frac8{N^2}(N^2+N) = 8 +\frac8N \] The second term \(\frac8N\) tends to zero as \(N\) approaches \(\infty\), so \[ A = \lim\limits_{ N \rightarrow \infty} R_N = \lim\limits_{ N \rightarrow \infty} \left(8 +\frac8N\right) = 8 \] As expected, this limit yields the same value as the formula \(\tfrac12bh\).
• (c) The left-endpoint approximation is similar, but the sum begins at \(j=0\) and ends at \(j= N-1\): \[ L_N = \frac{16}{N^2} \sum\limits_{j=0}^{N-1} j= \frac{16}{N^2} \sum\limits_{j=1}^{N-1} j = \frac{16}{N^2} \left(\frac{(N-1)N}2\right) = 8 - \frac8N \tag{6} \]

  In Eq. (6), we apply the formula \[ \sum\limits_{j=1}^N j = \frac{N(N+1)}2 \] with \(N-1\) in place of \(N\): \[ \sum\limits_{j=1}^{N-1} j = \frac{(N-1)N}2 \]

  Note in the second step that we replaced the sum beginning at \(j=0\) with a sum beginning at \(j=1\). This is valid because the term for \(j=0\) is zero and may be dropped. Again, we find that \[ A = \lim\limits_{ N \rightarrow \infty} L_N = \lim\limits_{ N \rightarrow \infty} \left(8 - \frac8N\right) = 8 \]

EXAMPLE 4

Let \(A\) be the area under the graph of \(f(x) = 2x^2 - x + 3\) over \([2, 4]\) (Figure 5.12). Compute \(A\) as the limit \(\lim\limits_{N \rightarrow\infty } R_N\).

Figure 5.12: Area under the graph of \(f(x) = 2x^2 - x + 3\) over \([2, 4]\).

Solution

Step 1. Express \(R_N\) in terms of power sums.

In this case, \(\Delta x = \tfrac{4-2}N = \tfrac2N\) and \[ R_N = \Delta x \sum\limits_{j=1}^N f(a+j\Delta x) = \frac2N \sum\limits_{j=1}^N f\left(2+\frac{2j}N\right) \] Let’s use algebra to simplify the general term. Since \(f(x)=2x^2-x+3\), \[ \begin{aligned} f\left(2+\frac{2j}N\right) & = 2 \left(2+\frac{2j}N\right)^2 - \left(2+\frac{2j}N\right) + 3\\ & = 2\left(4+\frac{8j}N + \frac{4j^2}{N^2}\right) - \left(2+\frac{2j}N\right) + 3 \\ & = \frac8{N^2} j^2 + \frac{14}N j + 9 \end{aligned} \] Now we can express \(R_N\) in terms of power sums: \[ \begin{align} R_N & = \frac2N \sum\limits_{j=1}^N \left(\frac8{N^2} j^2 + \frac{14}N j + 9\right) \\ & = \frac2N \sum\limits_{j=1}^N\frac8{N^2} j^2 + \frac2N\sum\limits_{j=1}^N\frac{14}N j + \frac2N\sum\limits_{j=1}^N9\\ & = \frac{16}{N^3}\sum\limits_{j=1}^N j^2 + \frac{28}{N^2}\sum\limits_{j=1}^N j + \frac{18}{N}\sum\limits_{j=1}^N 1 \tag{7} \end{align} \]

Step 2. Use the formulas for the power sums.

Using formulas (3) and (4) for the power sums in Eq. (7), we obtain \[ \begin{aligned} R_N & =\frac{16}{N^3} \left(\frac{N^3}3 + \frac{N^2}2 + \frac N6 \right) + \frac{28}{N^2} \left( \frac{N^2}2 + \frac N2\right) + \frac{18}N (N)\\ & = \left(\frac{16}3 + \frac8N + \frac8{3N^2} \right) + \left( 14 + \frac{14}N\right) + 18 \\ & = \frac{112}3 + \frac{22}N + \frac8{3N^2} \end{aligned} \]

Step 3. Calculate the limit.

\[ A = \lim\limits_{N \rightarrow\infty } \left(\frac{112}3 + \frac{22}N + \frac8{3N^2}\right) = \frac{112}3. \]

REMINDERBy Eq. (4) \[ \sum\limits_{j=1}^N j^2 = \frac{N^3}3 + \frac{N^2}2 + \frac N6 \]

EXAMPLE 5

Prove that for all \(b > 0\), the area \(A\) under the graph of \(f(x) = x^2\) over \([0, b]\) is equal to \(\tfrac{b^3}3\), as indicated in Figure 5.13.

Figure 5.13

Solution We’ll compute with \(R_N\). We have \(\Delta x = \tfrac{b-0}{N} = \tfrac{b}{N}\) and \[ R_N = \Delta x \sum\limits_{j=1}^N f(0 + j\Delta x) = \frac bN \sum\limits_{j=1}^N \left(0+j\frac b N\right)^2 = \frac bN \sum\limits_{j=1}^N \left(j^2\frac{b^2}{N^2}\right) = \frac{b^3}{N^3}\sum\limits_{j=1}^N j^2 \] By the formula for the power sum recalled in the margin, \[ \begin{aligned} R_N & = \frac{b^3}{N^3}\left(\frac{N^3}3 + \frac{N^2}2 + \frac N6 \right) = \frac{b^3}3 + \frac{b^3}{2N} + \frac{b^3}{6N^2} \\ &=\lim\limits_{N \rightarrow \infty } \left(\frac{b^3}3 + \frac{b^3}{2N} + \frac{b^3}{6N^2} \right) = \frac{b^3}3 \end{aligned} \]

HISTORICAL PERSPECTIVE

Jacob Bernoulli (1654–1705)

We used the formulas for the \(k\)^th power sums for \(k = 1, 2, 3\). Do similar formulas exist for all powers \(k\)? This problem was studied in the seventeenth century and eventually solved around 1690 by the great Swiss mathematician Jacob Bernoulli. Of this discovery, he wrote

"With the help of [these formulas] it took me less than half of a quarter of an hour to find that the 10^th powers of the first 1000 numbers being added together will yield the sum

Bernoulli’s formula has the general form \[ \sum\limits_{j=1}^n j^k = \frac1{k+1}n^{k+1} + \frac12 n^k + \frac k{12}n^{k-1} + \cdots \] The dots indicate terms involving smaller powers of n whose coefficients are expressed in terms of the so-called Bernoulli numbers. For example, \[ \sum\limits_{j=1}^n j^4 = \frac15 n^5 + \frac12n^4 + \frac13n^3 - \frac1{30}n \] These formulas are available on most computer algebra systems.