Approximating Areas Under Curves and Riemann Sum Foundations

Foundations of Calculus and Approximating Area

• Calculus is built upon three foundational geometric and physical questions:
  • The behavior of points infinitely close to each other or infinitely far away, which establishes the concept of limits.
  • The definition of a tangent line to a curve or the instantaneous rate of change, which establishes the concept of derivatives.
  • The definition of the area of curved shapes and the accumulation of changing quantities over an interval, which establishes the concept of integration (explored in Chapter 5).
• Development strategy for integration:
  • Rather than computing areas under curved curves directly from the outset, approximations are constructed using simple geometric polygons (specifically rectangles).
  • The approximation is subsequently transformed into an exact equality by evaluating the limit as the number of approximating rectangles approaches infinity.
• Section 5.1 focuses entirely on these rectangular approximations.
• Rectangles are selected because their geometric area is straightforward to compute:
  • $\text{Area} = \text{base} \times \text{height}$
• Area under the curve:
  • Defined as the area of the region bounded between the graph of a function and the $x$-axis over a specified interval.
• Mechanics of basic rectangular approximation:
  • Rectangular approximations are not exact because rectangles inevitably overshoot (extending above the curve and including unwanted area) or undershoot (failing to cover the entirety of the area beneath the curve).
  • Over a partition of the domain, the total approximate area is the sum of the individual rectangular areas:
  • $\text{Total Area} = (\text{base}_1 \times \text{height}_1) + (\text{base}_2 \times \text{height}_2) + \dots + (\text{base}_n \times \text{height}_n)$
  • The base corresponds to the width of the interval along the $x$-axis ($\Delta x$), and the height corresponds to the $y$-coordinate of the function.

Example 1: Arbitrary Rectangular Approximation

• Approximation of the area under a curve using five distinct rectangles with varying base widths and heights:
  • Rectangle 1:
  • Base width: $2$
  • Height: $6$
  • Area: $2 \times 6 = 12$
  • Rectangle 2:
  • Spans from $x = 2$ to $x = 4$, giving a base width of $4 - 2 = 2$
  • Height: $5$
  • Area: $2 \times 5 = 10$
  • Rectangle 3:
  • Spans from $x = 4$ to $x = 8$, giving a base width of $8 - 4 = 4$
  • Height: $3$
  • Area: $4 \times 3 = 12$
  • Rectangle 4:
  • Spans a base width of $4$
  • Height: $5$
  • Area: $4 \times 5 = 20$
  • Rectangle 5:
  • Spans a base width of $4$
  • Height: $7$
  • Area: $4 \times 7 = 28$
• Total area computation:
  • $\text{Total Area} = 12 + 10 + 12 + 20 + 28 = 82\,\text{square units}$
• While base widths do not strictly need to be identical, setting all bases to an equal width significantly streamlines systematic algebraic computations.

Right Endpoint Approximation ($R_n$)

• Subdividing an interval $[a, b]$ into $n$ equal subintervals:
  • The uniform base width (denoted $\Delta x$) is computed by taking the total length of the interval and dividing by the desired number of rectangles $n$:
  • $\Delta x = \frac{b - a}{n}$
  • The interval $[a, b]$ is partitioned into $n$ subintervals of equal width: $[x_{i-1}, x_i]$ for $i = 1, 2, \dots, n$.
  • Endpoints are computed iteratively by shifting to the right from $a$ by multiples of $\Delta x$:
  • $x_0 = a$
  • $x_1 = a + \Delta x$
  • $x_2 = a + 2\,\Delta x$
  • $x_i = a + i\,\Delta x$
  • $x_n = a + n\,\Delta x = b$
• Height selection for the right endpoint approximation:
  • Rectangles are constructed such that their top-right corners intersect the curve.
  • For each subinterval $[x_{i-1}, x_i]$, the height of the $i$-th rectangle is determined by evaluating the function at the right endpoint $x_i$:
  • $\text{Height}_i = f(x_i)$
• Formula for the $n$-th Right Endpoint Approximation ($R_n$):
  • Factoring out the uniform base width $\Delta x$ yields:
  • $R_n = \Delta x \cdot [f(x_1) + f(x_2) + \dots + f(x_n)]$
• Behavior as $n$ increases:
  • Making rectangles thinner by increasing $n$ diminishes the magnitude of both undershooting and overshooting gaps.
  • As $n \to \infty$, the approximation converges to the true area under the curve.

Example 2: Approximating the Area Under $f(x) = x^2$ on $[1, 3]$

• Problem objective: Estimate the area under $f(x) = x^2$ on the interval $[1, 3]$ using right endpoint approximations with $n = 4$ and $n = 6$.
• Part (a): Four rectangles ($n = 4$):
  • Calculate the uniform base width $\Delta x$:
  • $\Delta x = \frac{3 - 1}{4} = \frac{2}{4} = \frac{1}{2}$
  • Identify the subinterval endpoints by starting at $a = 1$ and repeatedly adding $\Delta x = \frac{1}{2}$:
  • $x_0 = 1$
  • $x_1 = 1 + \frac{1}{2} = \frac{3}{2}$
  • $x_2 = \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2$
  • $x_3 = \frac{4}{2} + \frac{1}{2} = \frac{5}{2}$
  • $x_4 = \frac{5}{2} + \frac{1}{2} = \frac{6}{2} = 3$
  • Calculate rectangular heights using $f(x) = x^2$:
  • $f(x_1) = f\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$
  • $f(x_2) = f\left(\frac{4}{2}\right) = \left(\frac{4}{2}\right)^2 = \frac{16}{4}$
  • $f(x_3) = f\left(\frac{5}{2}\right) = \left(\frac{5}{2}\right)^2 = \frac{25}{4}$
  • $f(x_4) = f\left(\frac{6}{2}\right) = \left(\frac{6}{2}\right)^2 = \frac{36}{4}$
  • Compute $R_4$:
  • $R_4 = \Delta x \cdot [f(x_1) + f(x_2) + f(x_3) + f(x_4)]$
  • $R_4 = \frac{1}{2} \left[ \frac{9}{4} + \frac{16}{4} + \frac{25}{4} + \frac{36}{4} \right]$
  • Sum of numerators: $9 + 16 + 25 + 36 = 86$
  • $R_4 = \frac{1}{2} \left( \frac{86}{4} \right) = \frac{86}{8} = 10.75$
• Part (b): Six rectangles ($n = 6$):
  • Calculate the uniform base width $\Delta x$:
  • $\Delta x = \frac{3 - 1}{6} = \frac{2}{6} = \frac{1}{3}$
  • Identify the subinterval endpoints by starting at $a = 1$ and repeatedly adding $\Delta x = \frac{1}{3}$:
  • $x_0 = 1$
  • $x_1 = 1 + \frac{1}{3} = \frac{4}{3}$
  • $x_2 = \frac{4}{3} + \frac{1}{3} = \frac{5}{3}$
  • $x_3 = \frac{5}{3} + \frac{1}{3} = \frac{6}{3} = 2$
  • $x_4 = \frac{6}{3} + \frac{1}{3} = \frac{7}{3}$
  • $x_5 = \frac{7}{3} + \frac{1}{3} = \frac{8}{3}$
  • $x_6 = \frac{8}{3} + \frac{1}{3} = \frac{9}{3} = 3$
  • Calculate rectangular heights using $f(x) = x^2$:
  • $f(x_1) = \left(\frac{4}{3}\right)^2 = \frac{16}{9}$
  • $f(x_2) = \left(\frac{5}{3}\right)^2 = \frac{25}{9}$
  • $f(x_3) = \left(\frac{6}{3}\right)^2 = \frac{36}{9}$
  • $f(x_4) = \left(\frac{7}{3}\right)^2 = \frac{49}{9}$
  • $f(x_5) = \left(\frac{8}{3}\right)^2 = \frac{64}{9}$
  • $f(x_6) = \left(\frac{9}{3}\right)^2 = \frac{81}{9}$
  • Compute $R_6$:
  • $R_6 = \frac{1}{3} \left[ \frac{16}{9} + \frac{25}{9} + \frac{36}{9} + \frac{49}{9} + \frac{64}{9} + \frac{81}{9} \right]$
  • Sum of numerators: $16 + 25 + 36 + 49 + 64 + 81 = 271$
  • $R_6 = \frac{1}{3} \left( \frac{271}{9} \right) = \frac{271}{27} \approx 10.04$
• Analysis of results:
  • The approximation decreased from $10.75$ to $10.04$.
  • Because $f(x) = x^2$ is an increasing function on $[1, 3]$, right endpoint rectangles consistently overshoot the actual area under the graph.
  • Increasing $n$ from $4$ to $6$ reduces the overcounting margin, driving the estimate closer to the actual value.

Summation (Sigma) Notation

• Definition and structure:
  • Summation notation provides a concise mathematical shorthand for expressing long sums of indexed numbers $a_m, a_{m+1}, \dots, a_n$:
  • $\sum_{i=m}^n a_i = a_m + a_{m+1} + a_{m+2} + \dots + a_n$
  • Read aloud as: "the sum as $i$ goes from $m$ to $n$ of $a_i$".
• Components of the notation:
  • Capital Greek letter Sigma ($\Sigma$): Represents "sum" (originating from the English letter S).
  • Summation index ($i$): A "dummy variable" that keeps track of the current term. Any symbol can serve as the index.
  • Lower limit ($m$): The starting integer value substituted into the expression.
  • Upper limit ($n$): The stopping integer value substituted into the expression.
• Algorithmic analogy:
  • Sigma notation is equivalent to a programming for loop initialized at $i = m$, running up to $i = n$, incrementing by $1$ each iteration, and accumulating the term $a_i$ into a total sum.
• Illustrative evaluation example:
  • $\sum_{i=3}^5 (i^2 + i)$
  • Term for $i = 3$: $3^2 + 3 = 9 + 3 = 12$
  • Term for $i = 4$: $4^2 + 4 = 16 + 4 = 20$
  • Term for $i = 5$: $5^2 + 5 = 25 + 5 = 30$
  • Sum evaluation:
  • $\sum_{i=3}^5 (i^2 + i) = (3^2 + 3) + (4^2 + 4) + (5^2 + 5) = 12 + 20 + 30 = 62$
• Compact representation of Right Endpoint Approximation ($R_n$):
  • $R_n = \Delta x \sum_{i=1}^n f(x_i)$

Left Endpoint and Midpoint Approximations

• Left Endpoint Approximation ($L_n$):
  • Rectangles are constructed such that their top-left corners touch the graph of $f(x)$.
  • On each subinterval $[x_{i-1}, x_i]$, the height is evaluated at the left boundary $x_{i-1}$:
  • $\text{Height}_i = f(x_{i-1})$
  • Expressed using an index starting at $1$:
  • $L_n = \Delta x \sum_{i=1}^n f(x_{i-1})$
  • Expressed alternatively by re-indexing to start at $0$ and stop at $n - 1$:
  • $L_n = \Delta x \sum_{i=0}^{n-1} f(x_i)$
  • The index begins at $x_0 = a$ and stops at $x_{n-1}$, omitting the right boundary $x_n = b$.
• Midpoint Approximation ($M_n$):
  • Rectangles are constructed such that their heights match the value of the function at the exact midpoint of each subinterval.
  • The midpoint of subinterval $[x_{i-1}, x_i]$ is defined as the arithmetic average of the two endpoints:
  • $x_i^* = \frac{x_{i-1} + x_i}{2}$
  • The height of the $i$-th rectangle is:
  • $\text{Height}_i = f\left(\frac{x_{i-1} + x_i}{2}\right)$
  • Formula for $M_n$:
  • $M_n = \Delta x \sum_{i=1}^n f\left(\frac{x_{i-1} + x_i}{2}\right)$
• Accuracy comparison across approximations:
  • For a fixed number of rectangles $n$, the midpoint approximation $M_n$ typically yields a substantially more accurate approximation than either $L_n$ or $R_n$.
  • Because the midpoint sample lies in the interior of each subinterval, the rectangle overshoots the curve on one side and undershoots it on the other side. These two opposite errors partially cancel each other out across each subinterval.

Physical Application: Estimating Distance from Velocity Data

• Physical foundations:
  • At a constant speed $v$, distance traveled across elapsed time $t$ is given by:
  • $\text{Distance} = \text{speed} \times \text{time} = v \cdot t$
  • When velocity is non-constant, this product is no longer directly applicable over the entire duration.
  • Rectangular approximations operate on the premise that over sufficiently small time increments (subintervals), velocity is approximately constant.
  • Total distance is approximated by accumulating the estimated distances across all subintervals.
• Example 3 setup and data:
  • An object's velocity $v(t)$ was recorded at half-second intervals over the span $t \in [0, 3]$:
  • At $t = 0.0\,\text{s}$: $v(0.0) = 0\,\text{m/s}$
  • At $t = 0.5\,\text{s}$: $v(0.5) = 12\,\text{m/s}$
  • At $t = 1.0\,\text{s}$: $v(1.0) = 18\,\text{m/s}$
  • At $t = 1.5\,\text{s}$: $v(1.5) = 25\,\text{m/s}$
  • At $t = 2.0\,\text{s}$: $v(2.0) = 20\,\text{m/s}$
  • At $t = 2.5\,\text{s}$: $v(2.5) = 14\,\text{m/s}$
  • At $t = 3.0\,\text{s}$: $v(3.0) = 20\,\text{m/s}$
• Part (a): Right endpoint approximation with six rectangles ($R_6$):
  • Calculate time increment $\Delta t$ for $n = 6$:
  • $\Delta t = \frac{3 - 0}{6} = 0.5\,\text{s}$
  • Right endpoint sample times: $t_1 = 0.5$, $t_2 = 1.0$, $t_3 = 1.5$, $t_4 = 2.0$, $t_5 = 2.5$, $t_6 = 3.0$
  • Computation:
  • $R_6 = \Delta t \sum_{i=1}^6 v(t_i)$
  • $R_6 = 0.5 \cdot [v(0.5) + v(1.0) + v(1.5) + v(2.0) + v(2.5) + v(3.0)]$
  • $R_6 = 0.5 \cdot [12 + 18 + 25 + 20 + 14 + 20]$
  • $R_6 = 0.5 \cdot [109] = 54.5\,\text{meters}$
• Part (b): Left endpoint approximation with six rectangles ($L_6$):
  • The subinterval width remains $\Delta t = 0.5\,\text{s}$.
  • Left endpoint sample times: $t_0 = 0.0$, $t_1 = 0.5$, $t_2 = 1.0$, $t_3 = 1.5$, $t_4 = 2.0$, $t_5 = 2.5$
  • Computation:
  • $L_6 = \Delta t \sum_{i=0}^5 v(t_i)$
  • $L_6 = 0.5 \cdot [v(0.0) + v(0.5) + v(1.0) + v(1.5) + v(2.0) + v(2.5)]$
  • $L_6 = 0.5 \cdot [0 + 12 + 18 + 25 + 20 + 14]$
  • $L_6 = 0.5 \cdot [89] = 44.5\,\text{meters}$
  • With $L_6 = 44.5\,\text{m}$ and $R_6 = 54.5\,\text{m}$, one method undershoots and the other overshoots; the actual total distance is estimated to lie near the center of these bounds (around $50\,\text{m}$.
• Part (c): Midpoint approximation with three rectangles ($M_3$):
  • Because $n = 3$, $\Delta t$ must be recalculated:
  • $\Delta t = \frac{3 - 0}{3} = 1.0\,\text{s}$
  • Subintervals: $[0, 1]$, $[1, 2]$, and $[2, 3]$
  • Subinterval midpoints:
  • For $[0, 1]$: $\frac{0 + 1}{2} = 0.5\,\text{s}$
  • For $[1, 2]$: $\frac{1 + 2}{2} = 1.5\,\text{s}$
  • For $[2, 3]$: $\frac{2 + 3}{2} = 2.5\,\text{s}$
  • Computation:
  • $M_3 = 1.0 \cdot [v(0.5) + v(1.5) + v(2.5)]$
  • $M_3 = 1.0 \cdot [12 + 25 + 14] = 51\,\text{meters}$
  • Result analysis:
  • The value $51\,\text{meters}$ lands directly between $L_6$ ($44.5\,\text{m}$) and $R_6$ ($54.5\,\text{m}$), demonstrating that the midpoint approximation achieves superior balance and accuracy even with half the number of rectangles ($n = 3$ vs. $n = 6$).

Convergence and the Definition of Area Under a Curve

• Equivalence of approximations in the limit:
  • Turning discrete geometric approximations into exact values requires taking the limit as the number of subintervals approaches infinity ($n \to \infty$).
  • For well-behaved functions, the choice of evaluation point within each subinterval (right endpoint, left endpoint, or midpoint) becomes irrelevant in the infinite limit.
• Formal Theorem for Continuous Functions:
  • Let $f$ be a continuous function defined on the closed interval $[a, b]$.
  • There exists a unique real number $A$ such that:
  • $\lim_{n \to \infty} R_n = \lim_{n \to \infty} L_n = \lim_{n \to \infty} M_n = A$
  • If $f(x) \ge 0$ for all $x \in [a, b]$, then $A \ge 0$.
  • This common limit $A$ is formally defined as the exact area under the curve $y = f(x)$ on $[a, b]$.
• Theoretical role of continuity in convergence:
  • The width of each subinterval is $\Delta x = \frac{b - a}{n}$.
  • As $n \to \infty$, $\Delta x \to 0$, meaning the endpoints $x_{i-1}$, $x_i$, and the midpoint $\frac{x_{i-1} + x_i}{2}$ converge toward the exact same value.
  • By the fundamental definition of continuity, if input coordinates get arbitrarily close to one another, their corresponding function outputs $f(x_{i-1})$, $f(x_i)$, and $f\left(\frac{x_{i-1} + x_i}{2}\right)$ must also get arbitrarily close to one another.
  • Consequently, the evaluations across all three methods converge to the identical sum in the limit.

Questions & Discussion

• Summing an infinite collection of rectangles:
  • Question: How do we add up an infinite amount of rectangles?
  • Answer: By evaluating limits. The discrete sum with $n$ rectangles is evaluated under the limit process as $n \to \infty$.
• Upper bound in sigma notation:
  • Question: The $n$ on the top is basically the number we stop at?
  • Answer: Yes. The variable $m$ at the base of the sigma denotes the starting index, and the variable $n$ at the top denotes the stopping index. It is important to emphasize that these limits are not the numbers being summed; they are indices that specify which evaluated terms enter the summation.
• Dummy variables and indices:
  • Question: Are the $i$ and the $n$ like normal variables?
  • Answer: The variable $i$ is a dummy variable/index that keeps track of the progression through the sum, and any letter can be substituted in its place. The variable $n$ represents the total number of rectangles into which the interval is partitioned. Because right endpoint approximations evaluate after shifting over by $\Delta x$, they initialize at $i = 1$.
• Starting index for left endpoint approximations:
  • Question: And left endpoint thing was a thing when we start with zero?
  • Answer: Yes. The left endpoint approximation begins evaluating at the far left boundary $x_0 = a$, which corresponds to starting the summation index at $i = 0$ and terminating at $n - 1$.
• Behavior when functions reach zero or negative values:
  • Question: What if our line goes to zero?
  • Answer: That condition is addressed in subsequent sections. For introductory approximations, the function is assumed to satisfy $f(x) \ge 0$ to keep the geometric concept of area straightforward.
• Derivation of evaluation points in midpoint approximation:
  • Question: How do we know to derive the equation for $t$? Like, where am I getting these numbers from?… The line under that $M_3$ equals summation…
  • Answer: Those values are obtained because the midpoint approximation requires evaluating the function at the center of each subinterval, which is the average of successive endpoints. For subintervals $[0, 1]$, $[1, 2]$, and $[2, 3]$, the midpoints are $\frac{0 + 1}{2} = 0.5$, $\frac{1 + 2}{2} = 1.5$, and $\frac{2 + 3}{2} = 2.5$. These times are then referenced directly in the table to retrieve their respective velocities.