Approximating Areas with | AP Calc BC Unit 6 Study Guide

The Core Idea: Approximating Areas with Riemann Sums

The definite integral is a fundamental concept in calculus that represents the net area under a curve over a given interval. While we will learn methods to calculate this area exactly, it is often necessary or practical to find an approximation. The core idea of this topic is to approximate the definite integral by dividing the area into simpler, more manageable geometric shapes—specifically, rectangles or trapezoids.

A Riemann sum is the sum of the areas of these approximating shapes. By summing the areas of a finite number of rectangles whose heights are determined by the function's value at a specific point in each subinterval (left endpoint, right endpoint, or midpoint), we can generate an estimate of the total area. Similarly, using trapezoids can provide another approximation. These methods are powerful because they can be applied to functions represented in various ways: graphically, numerically in a table, or by an explicit formula.

Key Formulas

The definite integral $\int_{a}^{b} f (x) d x$ can be approximated by dividing the interval $[a, b]$ into $n$ subintervals. Let the width of each subinterval be $Δ x = \frac{b - a}{n}$ . The endpoints of the subintervals are $x_{0} = a, x_{1}, x_{2}, \dots, x_{n} = b$ .

Left Riemann Sum

The height of each rectangle is determined by the function's value at the left endpoint of each subinterval.

$L_{n} = i = 0 \sum n - 1 f (x_{i}) Δ x = [f (x_{0}) + f (x_{1}) + \dots + f (x_{n - 1})] Δ x$

Right Riemann Sum

The height of each rectangle is determined by the function's value at the right endpoint of each subinterval.

$R_{n} = i = 1 \sum n f (x_{i}) Δ x = [f (x_{1}) + f (x_{2}) + \dots + f (x_{n})] Δ x$

Midpoint Riemann Sum

The height of each rectangle is determined by the function's value at the midpoint, $m_{i} = \frac{x _{i - 1} + x _{i}}{2}$ , of each subinterval.

$M_{n} = i = 1 \sum n f (m_{i}) Δ x = [f (m_{1}) + f (m_{2}) + \dots + f (m_{n})] Δ x$

Trapezoidal Sum

This method approximates the area using trapezoids instead of rectangles. The area of each trapezoid is $\frac{1}{2} (base_{1} + base_{2}) \times height$ , where the bases are the parallel vertical sides ( $f (x_{i - 1})$ and $f (x_{i})$ ) and the height is the width of the subinterval ( $Δ x$ ).

$T_{n} = i = 1 \sum n \frac{f ( x _{i - 1} ) + f ( x _{i} )}{2} Δ x$

For subintervals of equal width, this simplifies to:

$T_{n} = \frac{Δ x}{2} [f (x_{0}) + 2 f (x_{1}) + 2 f (x_{2}) + \dots + 2 f (x_{n - 1}) + f (x_{n})]$

Understanding the Methods

The choice of approximation method depends on which point in each subinterval is used to determine the height of the approximating shape. This choice is the fundamental difference between the various sums.

Left vs. Right Riemann Sums: For a left sum, the top-left corner of each rectangle touches the curve. For a right sum, the top-right corner touches the curve. If the function is increasing on the interval, the left sum will be an underestimate of the true area, while the right sum will be an overestimate. The reverse is true for a decreasing function.
Midpoint Riemann Sum: This method often provides a better approximation than left or right sums because the parts of the rectangle that are above the curve often cancel out the parts that are below the curve within each subinterval. The height is determined by the function value at the exact center of each subinterval's base.
Trapezoidal Sum: By using a slanted top for each shape, the trapezoidal sum conforms more closely to the curve than a rectangle with a flat top. It is equivalent to averaging the left and right Riemann sums for subintervals of equal width: $T_{n} = \frac{L _{n} + R _{n}}{2}$ .

A critical detail, especially when working with data from a table, is that the subintervals may not have equal widths. In such cases, the width of each individual rectangle or trapezoid ( $Δ x_{i} = x_{i} - x_{i - 1}$ ) must be calculated and used for that specific shape's area calculation. The simplified formulas that factor out a common $Δ x$ do not apply.

Core Concepts & Rules

A Riemann sum is a method for approximating a definite integral by summing the areas of a finite number of rectangles.
The primary approximation methods are the left Riemann sum, right Riemann sum, midpoint Riemann sum, and the trapezoidal sum.
The specific method is defined by the point chosen within each subinterval to determine the height of the rectangle (left endpoint, right endpoint, or midpoint) or by the use of trapezoids.
These approximation methods are versatile and can be applied to functions presented as an equation, a graph, or a table of numerical values.
When approximating from a table of values, be aware that the width of the subintervals ( $Δ x$ ) may not be constant.

Step-by-Step Example 1: Approximating from a Function

Approximate $\int_{1}^{3} (x^{2} + 1) d x$ using a right Riemann sum with $n = 4$ subintervals.

Step 1: Determine the width of each subinterval, $Δ x$ .

The interval is $[a, b] = [1, 3]$ and $n = 4$ .

$Δ x = \frac{b - a}{n} = \frac{3 - 1}{4} = \frac{2}{4} = 0.5$

Step 2: Determine the $x$ -values for the endpoints of the subintervals.

Starting at $x_{0} = 1$ , we add $Δ x = 0.5$ repeatedly.

$x_{0} = 1$
$x_{1} = 1 + 0.5 = 1.5$
$x_{2} = 1.5 + 0.5 = 2.0$
$x_{3} = 2.0 + 0.5 = 2.5$
$x_{4} = 2.5 + 0.5 = 3.0$

Step 3: Identify the $x$ -values to be used for the heights.

For a right Riemann sum, we use the right endpoints of each of the 4 subintervals: $x_{1}, x_{2}, x_{3}, x_{4}$ .

The values are: $1.5, 2.0, 2.5, 3.0$ .

Step 4: Calculate the function value (height) at each of these $x$ -values.

Let $f (x) = x^{2} + 1$ .

$f (1.5) = (1.5)^{2} + 1 = 2.25 + 1 = 3.25$
$f (2.0) = (2.0)^{2} + 1 = 4 + 1 = 5.0$
$f (2.5) = (2.5)^{2} + 1 = 6.25 + 1 = 7.25$
$f (3.0) = (3.0)^{2} + 1 = 9 + 1 = 10.0$

Step 5: Calculate the sum of the areas of the rectangles.

The area is the sum of $(height) \times (width)$ .

$R_{4} = [f (1.5) + f (2.0) + f (2.5) + f (3.0)] Δ x$

$R_{4} = [3.25 + 5.0 + 7.25 + 10.0] \times 0.5$

$R_{4} = [25.5] \times 0.5 = 12.75$

The right Riemann sum approximation is $12.75$ .

Step-by-Step Example 2: Approximating from a Table of Values

A particle's velocity $v (t)$ (in meters/sec) is recorded at various times $t$ (in seconds) as shown in the table below. Use a trapezoidal sum with the three subintervals indicated by the data to approximate the total distance traveled, $\int_{0}^{10} v (t) d t$ .

$t$ (sec)	0	4	7	10
$v (t)$ (m/s)	5	8	12	11

Step 1: Identify the subintervals and their widths.

The subintervals are determined by the $t$ -values in the table. Note that the widths are not equal.

Subinterval 1: $[0, 4]$ . Width $Δ t_{1} = 4 - 0 = 4$ .
Subinterval 2: $[4, 7]$ . Width $Δ t_{2} = 7 - 4 = 3$ .
Subinterval 3: $[7, 10]$ . Width $Δ t_{3} = 10 - 7 = 3$ .

Step 2: Calculate the area of the trapezoid for each subinterval.

The formula for the area of a trapezoid is $\frac{1}{2} (b_{1} + b_{2}) h$ . Here, the "bases" are the vertical function values ( $v (t)$ ) and the "height" is the horizontal width of the subinterval ( $Δ t$ ).

Area 1 (for $[0, 4]$ ):
$\frac{1}{2} (v (0) + v (4)) Δ t_{1} = \frac{1}{2} (5 + 8) \times 4 = \frac{1}{2} (13) \times 4 = 26$
Area 2 (for $[4, 7]$ ):
$\frac{1}{2} (v (4) + v (7)) Δ t_{2} = \frac{1}{2} (8 + 12) \times 3 = \frac{1}{2} (20) \times 3 = 30$
Area 3 (for $[7, 10]$ ):
$\frac{1}{2} (v (7) + v (10)) Δ t_{3} = \frac{1}{2} (12 + 11) \times 3 = \frac{1}{2} (23) \times 3 = 34.5$

Step 3: Sum the areas of the trapezoids.

Total Approximation = Area 1 + Area 2 + Area 3

$\int_{0}^{10} v (t) d t \approx 26 + 30 + 34.5 = 90.5$

The approximate distance traveled is $90.5$ meters.

Using Your Calculator

The calculation of Riemann and trapezoidal sums is fundamentally an arithmetic process that is expected to be done by hand, especially when data is provided in a table. A calculator's primary role in this topic is not to perform the approximation but to speed up the arithmetic or to find the exact value of the definite integral for comparison.

Checking an Answer for a Function:

If you have approximated an integral like $\int_{1}^{3} (x^{2} + 1) d x$ , you can use your calculator to find the exact value and see if your approximation is reasonable.

On a TI-84 style calculator, use the numerical integration function, typically found under the [MATH] menu.
The syntax is fnInt(function, variable, lower_bound, upper_bound).
For the example above: $f n I n t (X^{2} + 1, X, 1, 3)$ would yield approximately $10.667$ . Our right-sum approximation of $12.75$ is in a reasonable range for an estimate with only 4 subintervals.

Speeding up Arithmetic:

For a function with many subintervals, you can use lists to store the $x$ -values and $f (x)$ -values and then use list operations to perform the sum. However, setting up the sum correctly by hand is the key skill being assessed.

AP Exam Quick Hit

Common Question Types

Approximation from a Table: This is the most frequent format. You will be given a table of values for a function (e.g., rate of change) and asked to approximate the definite integral (e.g., total accumulation) using a left sum, right sum, or trapezoidal sum. The subintervals are often of unequal width.
- Example: "Use a trapezoidal sum with 4 subintervals given by the data in the table to approximate the amount of water that leaked from a tank."
Approximation from a Graph: You will be shown a graph of a function $f$ and asked to use a specified number of subintervals to approximate $\int_{a}^{b} f (x) d x$ . This requires you to read the function values (the heights) directly from the graph at the required endpoints.
- Example: "Using a midpoint Riemann sum with 3 equal subintervals, approximate the value of $\int_{0}^{6} f (x) d x$ for the function $f$ shown in the graph."

Common Mistakes

Using the Wrong Number of Function Values: For $n$ subintervals, a left or right Riemann sum uses $n$ function values, but the trapezoidal sum uses $n + 1$ function values.
Incorrectly Identifying Endpoints: For a left sum on $[x_{0}, x_{n}]$ , using $f (x_{1}), \dots, f (x_{n})$ instead of the correct $f (x_{0}), \dots, f (x_{n - 1})$ . The reverse is a common mistake for right sums.
Ignoring Unequal Subintervals: When working from a table, automatically factoring out a single $Δ x$ value when the widths of the subintervals are different. You must calculate the area of each shape individually using its specific width.
Trapezoidal Formula Error: Forgetting the $\frac{1}{2}$ factor in the trapezoid area formula. For the simplified formula with equal subintervals, a common error is forgetting to apply the coefficient of 2 to all interior function values.
Confusing Midpoint and Trapezoidal Sums: Calculating the height for a midpoint sum by averaging the function values at the endpoints ( $\frac{f ( a ) + f ( b )}{2}$ ) instead of evaluating the function at the midpoint of the x-values ( $f (\frac{a + b}{2})$ ).

Approximating Areas with Riemann Sums - AP Calculus BC Study Guide