5.1 Approximating and Computing Area

Why might we be interested in the area under a graph? Consider an object moving in a straight line with constant velocity v (assumed positive). The distance traveled over a time interval [t₁, t₂] is equal to vΔt where Δt = (t₂ − t₁) is the time elapsed. This is the well-known formula

Because v is constant, the graph of velocity is a horizontal line (Figure 1) and vΔt is equal to the area of the rectangular region under the graph of velocity over [t₁, t₂]. So we can write Eq. (1) as

There is, however, an important difference between these two equations: Eq. (1) makes sense only if velocity υ is constant whereas Eq. (2) is correct even if the velocity changes with time (we will prove this in Section 5.5). Thus, the advantage of expressing distance traveled as an area is that it enables us to deal with much more general types of motion.

To see why Eq. (2) might be true in general, let’s consider the case where velocity changes over time but is constant on intervals. In other words, we assume that the object’s velocity changes abruptly from one interval to the next as in Figure 2. The distance traveled over each interval is equal to the area of the rectangle above that interval, so the total distance traveled is the sum of the areas of the rectangles. In Figure 2,

Figure 5.1: The rectangle has area vΔt, which is equal to the distance traveled.

Figure 5.2: Distance traveled equals the sum of the areas of the rectangles.

Our strategy when velocity changes continuously (Figure 3) is to approximate the area under the graph by sums of areas of rectangles and then pass to a limit. This idea leads to the concept of an integral.

Figure 5.3: Distance traveled is equal to the area under the graph. It is approximated by the sum of the areas of the rectangles.

Approximating Area by Rectangles

Our goal is to compute the area under the graph of a function f(x). In this section, we assume that f(x) is continuous and positive, so that the graph of f(x) lies above the x-axis (Figure 4). The first step is to approximate the area using rectangles.

Figure 5.4

To begin, choose a whole number N and divide [a, b] into N subintervals of equal width, as in Figure 4(A). The full interval [a, b] has width b − a, so each subinterval has width Δx = (b − a)/N. The right endpoints of the subintervals are

Note that the last right endpoint is b because a + NΔx = a + N((b − a)/N) = b. Next, as in Figure 4(B), construct, above each subinterval, a rectangle whose height is the value of f(x) at the right endpoint of the subinterval.

The sum of the areas of these rectangles provides an approximation to the area under the graph. The first rectangle has base Δx and height f(a + Δx), so its area is f(a + Δx)Δx. Similarly, the second rectangle has height f(a + 2Δx) and area f(a + 2Δx) Δx, etc. The sum of the areas of the rectangles is denoted R_N and is called the Nth right-endpoint approximation:

Factoring out Δx, we obtain the formula

In words: R_N is equal to Δx times the sum of the function values at the right endpoints of the subintervals.

EXAMPLE 1

Calculate R₄ and R₆ for f(x) = x² on the interval [1, 3].

Solution

Step 1. Determine Δx and the right endpoints.

To calculate R₄, divide [1, 3] into four subintervals of width . The right endpoints are the numbers for j = 1, 2, 3, 4. They are spaced at intervals of beginning at , so, as we see in Figure 5(A), the right endpoints are , , , .

Figure 5.5

Step 2. Calculate Δx times the sum of function values.

R₄ is Δx times the sum of the function values at the right endpoints:

R₆ is similar: , and the right endpoints are spaced at intervals of beginning at and ending at 3, as in Figure 5(B). Thus,

Summation Notation

Summation notation is a standard notation for writing sums in compact form. The sum of numbers a_m,…, a_n (m ≤ n) is denoted

The Greek letter Σ (capital sigma) stands for “sum,” and the notation tells us to start the summation at j = m and end it at j = n. For example,

In this summation, the j th term is a_j = j². We refer to j² as the general term. The letter j is called the summation index. It is also referred to as a dummy variable because any other letter can be used instead. For example,

The usual commutative, associative, and distributive laws of addition give us the following rules for manipulating summations.

For example,

is equal to

Linearity can be used to write a single summation as a sum of several summations. For example,

It is convenient to use summation notation when working with area approximations. For example, R_N is a sum with general term f(a + jΔx):

The summation extends from j = 1 to j = N, so we can write R_N concisely as

We shall make use of two other rectangular approximations to area: the left-endpoint and the midpoint approximations. Divide [a, b] into N subintervals as before. In the left-endpoint approximation L_N, the heights of the rectangles are the values of f(x) at the left endpoints [Figure 6(A)]. These left endpoints are

Figure 5.6

and the sum of the areas of the left-endpoint rectangles is

Note that both R_N and L_N have general term f(a + jΔx), but the sum for L_N runs from j = 0 to j = N − 1 rather than from j = 1 to j = N:

In the midpoint approximation M_N, the heights of the rectangles are the values of f(x) at the midpoints of the subintervals rather than at the endpoints. As we see in Figure 6(B), the midpoints are

The sum of the areas of the midpoint rectangles is

In summation notation,

EXAMPLE 2

Calculate R₆, L₆, and M₆ for f(x) = x⁻¹ on [2, 4].

Solution In this case, . The general term in the summation for R₆ and L₆ is

Therefore (Figure 7),

Figure 5.7: L₆ and R₆ for f(x) = x⁻¹ on [2, 4].

In L₆, the sum begins at j = 0 and ends at j = 5:

The general term in M₆ is

Summing up from j = 1 to 6, we obtain (Figure 8)

Figure 5.8: M₆ for f(x) = x⁻¹ on [2, 4].

GRAPHICAL INSIGHT Monotonic Functions

Observe in Figure 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 R₆ and L₆, and so, according to the previous example, 0.65 ≤ A ≤ 0.74. More generally, when f(x) is monotonic (increasing or decreasing), the exact area lies between R_N and L_N (Figure 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) increasing L_N ≤ area under graph ≤ R_N
• f(x) decreasing R_N ≤ area under graph ≤ L_N

Computing Area as the Limit of Approximations

Figure 10 shows several right-endpoint approximations. Notice that the error, corresponding to the yellow region above the graph, gets smaller as the number of rectangles increases. In fact, it appears that we can make the error as small as we please by taking the number N of rectangles large enough. If so, it makes sense to consider the limit as N → ∞, as this should give us the exact area under the curve. The next theorem guarantees that the limit exists (see Theorem 8 in Appendix D for a proof and Exercise 87 for a special case).

Figure 5.10: The error decreases as we use more rectangles.

The next three examples illustrate Theorem 1 using formulas for power sums. The kth power sum is the sum of the kth powers of the first N integers. We shall use the power sum formulas for k = 1, 2, 3.

For example, by Eq. (4),

As a first illustration, we compute the area of a right triangle “the hard way.”

EXAMPLE 3

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

• (a) Using geometry
• (b)
• (c)

Solution The region under the graph is a right triangle with base b = 4 and height h = 4 (Figure 11).

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

• (a) By geometry, .
• (b) We compute this area again as a limit. Since Δx = (b − a)/N = 4/N and f(x) = x,
  

  REMINDER

  

  In the last equality, we factored out 4/N from the sum. This is valid because 4/N is a constant that does not depend on j. Now use formula (3):
  
  The second term 8/N tends to zero as N approaches ∞, so
  
  As expected, this limit yields the same value as the formula .
• (c) The left-endpoint approximation is similar, but the sum begins at j = 0 and ends at j = N − 1:
  
  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 .

In the next example, we compute the area under a curved graph. Unlike the previous example, it is not possible to compute this area directly using geometry.

EXAMPLE 4

Let A be the area under the graph of f(x) = 2x² − x + 3 over [2, 4] (Figure 12). Compute A as the limit .

Figure 5.12: Area under the graph of f(x) = 2x² − x + 3 over [2, 4].

Solution

Step 1. Express R_N in terms of power sums.

In this case, Δx = (4 − 2)/N = 2/N and

Let’s use algebra to simplify the general term. Since f(x) = 2x² − x + 3,

Now we can express R_N in terms of power sums:

Step 2. Use the formulas for the power sums.

Using formulas (3) and (4) for the power sums in Eq. (7), we obtain

Step 3. Calculate the limit.

EXAMPLE 5

Prove that for all b > 0, the area A under the graph of f(x) = x² over [0, b] is equal to b³/3, as indicated in Figure 13.

Figure 5.13

Solution We’ll compute with R_N. We have Δx = (b − 0)/N = b/N and

By the formula for the power sum recalled in the margin,

The area under the graph of any polynomial can be calculated using power sum formulas as in the examples above. For other functions, the limit defining the area may be hard or impossible to evaluate directly. Consider f(x) = sin x on the interval . In this case (Figure 14), Δx = (3π/4 − π/4)/N = π/(2N) and the area A is

Figure 5.14: The area of this region is more difficult to compute as a limit of endpoint approximations.

With some work, we can show that the limit is equal to . However, in Section 5.3 we will see that it is much easier to apply the Fundamental Theorem of Calculus, which reduces area computations to the problem of finding antiderivatives.

HISTORICAL PERSPECTIVE

Jacob Bernoulli (1654–1705)

We used the formulas for the kth 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 10th powers of the first 1000 numbers being added together will yield the sum

91409924241424243424241924242500

Bernoulli’s formula has the general form

The dots indicate terms involving smaller powers of n whose coefficients are expressed in terms of the so-called Bernoulli numbers. For example,

These formulas are available on most computer algebra systems.

5.1.1 Summary

• Approximations to the area under the graph of f(x) over [a, b] :
  

  Power Sums

  
• If f(x) is continuous on [a, b], then the endpoint and midpoint approximations approach one and the same limit L:

  

• If f(x) ≥ 0 on [a, b], we take L as the definition of the area under the graph of y = f(x) over [a, b].