Areas, Riemann Sums, and Definite Integrals

Geometric Foundations of Area and the Definite Integral

• Core Objective of Integration: Calculus 2 conceptualizes the integral as a systematic method for calculating areas by partitioning a region into increasingly narrow slices, summing the individual areas of those slices, and evaluating the limit as the slice count approaches infinity.

• Assumptions on Functions: For a curve defined by $y = f(x)$, initial geometric development assumes $f(x) \ge 0$. However, the analytical framework and definitions remain fully valid and applicable when $f(x) < 0$.

• Failure of Elementary Geometry: Because typical curves are curved rather than straight, exact areas enclosed between the curve and the $x$-axis cannot generally be calculated using elementary geometric area formulas.

• Geometric Analogy: The process parallels the classical method of estimating the area of a circle by inscribing skinnier and skinnier triangles. For the area bounded between a curve $y = f(x)$ and the $x$-axis across an interval $[a, b]$, approximating rectangles are used instead of triangles, where the function value $f(x)$ determines the height of a rectangle of infinitesimal width $\Delta x$.

Partitions and Riemann Sums

• Definition of a Partition: A set of points $P = \{x_i\}_{i=0}^n$ such that   $a = x_0 < x_1 < x_2 < \dots < x_{n-1} < x_n = b$   which divides the closed interval $[a, b]$ into $n$ subintervals of the form $[x_0, x_1], [x_1, x_2], \dots, [x_{n-1}, x_n]$ is defined as a partition of $[a, b]$.

• Definition of a Regular Partition: If every subinterval in the partition $P$ has an identical uniform width $\Delta x$, then $P$ is termed a regular partition of $[a, b]$, where:   $\Delta x = \frac{b - a}{n}$

• Definition of a Riemann Sum: Let $f(x)$ be a function defined on the closed interval $[a, b]$, and let $P$ be a regular partition of $[a, b]$ with $n$ subintervals of width $\Delta x = \frac{b - a}{n}$. For each index $i$, let $x_i^*$ denote an arbitrary sample point within the subinterval $[x_{i-1}, x_i]$. The Riemann sum is defined as:   $\sum_{i=1}^n f(x_i^*)\Delta x = f(x_1^*)\Delta x + f(x_2^*)\Delta x + \dots + f(x_n^*)\Delta x$

• Definition of Upper Sum: When each sample point $x_i^*$ is specifically chosen such that $f(x_i^*)$ is the absolute maximum value of $f(x)$ across the subinterval $[x_{i-1}, x_i]$, the resulting Riemann sum is designated as an upper sum.

• Definition of Lower Sum: When each sample point $x_i^*$ is chosen such that $f(x_i^*)$ is the absolute minimum value of $f(x)$ across the subinterval $[x_{i-1}, x_i]$, the resulting Riemann sum is designated as a lower sum.

• Common Choices for Sample Points ($x_i^*$):

  • Left-hand Sum: Sample point chosen at the left endpoint of each subinterval: $x_i^* = x_{i-1}$

  • Right-hand Sum: Sample point chosen at the right endpoint of each subinterval: $x_i^* = x_i$

  • Midpoint Sum: Sample point chosen at the midpoint of each subinterval: $x_i^* = \frac{x_{i-1} + x_i}{2}$

• Inequality Bounds and the Squeeze Theorem:

  • For any arbitrary choice of sample point $x_i^* \in [x_{i-1}, x_i]$, the function value satisfies $\min_{[x_{i-1}, x_i]} f(x) \le f(x_i^*) \le \max_{[x_{i-1}, x_i]} f(x)$. Consequently:     $\text{lower sum} \le \sum_{i=1}^n f(x_i^*)\Delta x \le \text{upper sum}$

  • The true area between the curve $y = f(x)$ and the $x$-axis is strictly bounded:     $\text{lower sum} < \text{area} < \text{upper sum}$

  • By the Squeeze Theorem, if the limit of the lower sum equals the limit of the upper sum as the subintervals shrink ($n \to \infty$), that shared limit is precisely equal to the exact area.

Finite Riemann Sum Computations

• Example Problem: Compute the lower and upper sums for $f(x) = x$ on the interval $[0, 3]$ using $n = 6$ equal subintervals.

• Partition Geometry:

  • Width of each subinterval:     $\Delta x = \frac{3 - 0}{6} = \frac{1}{2}$

  • Subinterval endpoints are given by:     $x_i = \frac{i}{2}$

  • Explicit subintervals: $[0, 0.5]$, $[0.5, 1]$, $[1, 1.5]$, $[1.5, 2]$, $[2, 2.5]$, and $[2.5, 3]$.

  • Because $f(x) = x$ is strictly increasing, the minimum value on each subinterval $[x_{i-1}, x_i]$ occurs at the left endpoint $x_{i-1}$, while the maximum value occurs at the right endpoint $x_i$.

• Lower Sum Calculation ($n = 6$):   $\sum_{i=1}^6 f(x_{i-1})\Delta x = \sum_{i=1}^6 f\left(\frac{i - 1}{2}\right)\frac{1}{2}$   $= \sum_{i=1}^6 \left(\frac{i - 1}{2}\right)\left(\frac{1}{2}\right)$   $= \frac{1}{2}\left(\frac{0}{2} + \frac{1}{2} + \frac{2}{2} + \frac{3}{2} + \frac{4}{2} + \frac{5}{2}\right)$   $= \frac{1}{4}(0 + 1 + 2 + 3 + 4 + 5)$   $= \frac{15}{4}$

• Upper Sum Calculation ($n = 6$):   $\sum_{i=1}^6 f(x_i)\Delta x = \sum_{i=1}^6 f\left(\frac{i}{2}\right)\frac{1}{2}$   $= \sum_{i=1}^6 \left(\frac{i}{2}\right)\left(\frac{1}{2}\right)$   $= \frac{1}{2}\left(\frac{1}{2} + \frac{2}{2} + \frac{3}{2} + \frac{4}{2} + \frac{5}{2} + \frac{6}{2}\right)$   $= \frac{1}{4}(1 + 2 + 3 + 4 + 5 + 6)$   $= \frac{21}{4}$

Limits of Upper and Lower Sums as $n \to \infty$

• Example Problem: Compute the limit as $n \to \infty$ of both the lower sum and upper sum for $f(x) = x$ on the interval $[0, 3]$.

• General Partition Setup for $n$ Subintervals:

  • Width of each subinterval:     $\Delta x = \frac{3 - 0}{n} = \frac{3}{n}$

  • Endpoints of the regular partition:     $x_i = \frac{3}{n} \cdot i = \frac{3i}{n}$

• Gauss's Summation Identity:   $\sum_{i=1}^n i = \frac{n(n + 1)}{2}$

• Limit of the Lower Sum:   $\sum_{i=1}^n f(x_{i-1})\Delta x = \sum_{i=1}^n \frac{3(i - 1)}{n} \cdot \frac{3}{n}$   $= \frac{9}{n^2}\sum_{i=1}^n (i - 1)$   $= \frac{9}{n^2}\left(-n + \sum_{i=1}^n i\right)$   $= -\frac{9}{n} + \frac{9}{n^2}\left(\frac{n(n + 1)}{2}\right)$   $= -9 \cdot \frac{1}{n} + \frac{9(n + 1)}{2n}$   $= -9 \cdot \frac{1}{n} + \frac{9}{2} + \frac{9}{2} \cdot \frac{1}{n}$

  • Evaluating the limit as $n \to \infty$ (where $\frac{1}{n} \to 0$):     $\lim_{n \to \infty} \sum_{i=1}^n f(x_{i-1})\Delta x = \frac{9}{2}$

• Limit of the Upper Sum:   $\sum_{i=1}^n f(x_i)\Delta x = \sum_{i=1}^n \frac{3i}{n} \cdot \frac{3}{n}$   $= \frac{9}{n^2}\sum_{i=1}^n i$   $= \frac{9}{n^2}\left(\frac{n(n + 1)}{2}\right)$   $= \frac{9(n + 1)}{2n}$   $= \frac{9}{2} + \frac{9}{2} \cdot \frac{1}{n}$

  • Evaluating the limit as $n \to \infty$ (where $\frac{1}{n} \to 0$):     $\lim_{n \to \infty} \sum_{i=1}^n f(x_i)\Delta x = \frac{9}{2}$

• Geometric Verification: The region enclosed by $f(x) = x$, the $x$-axis, and $x = 3$ forms a right triangle with base $3$ and height $3$. Standard geometry verifies the area:   $\text{Area} = \frac{1}{2} \cdot 3 \cdot 3 = \frac{9}{2}$

The Definite Integral and Signed Area

• Definition of the Definite Integral: The definite integral of $f$ from $a$ to $b$ is defined by:   $\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*)\Delta x$   provided that this limit exists, in which case $f(x)$ is said to be integrable on $[a, b]$.

• Signed Area Concept: The definite integral represents the net signed area bounded between the curve $y = f(x)$ and the $x$-axis across $[a, b]$:

  • Regions where $f(x) > 0$ (above the $x$-axis) yield positive area contributions.

  • Regions where $f(x) < 0$ (below the $x$-axis) yield negative area contributions.

• Integrability Theorem: If $f(x)$ is continuous on the closed interval $[a, b]$, then $f$ is integrable on $[a, b]$. Furthermore, continuity guarantees that any valid choice of Riemann sum (upper, lower, midpoint, left-endpoint, right-endpoint) converges to the exact value of the definite integral.

• Geometric Computation Example 1: $\int_0^6 (2x - 4)\,dx$:

  • The linear function $y = 2x - 4$ has slope $2$, $y$-intercept at $(0, -4)$, and an $x$-intercept at $(2, 0)$.

  • The area is partitioned into two distinct triangles across the interval $[0, 6]$:

  • Region below $x$-axis ($[0, 2]$): Since $f(x) = 2x - 4 \le 0$, this region contributes a negative signed area. The triangle has base $2$ (from $0$ to $2$) and height $4$ (magnitude at $x = 0$):       $\text{Contribution} = -\left(\frac{1}{2} \cdot 2 \cdot 4\right) = -4$

  • Region above $x$-axis ($[2, 6]$): Since $f(x) \ge 0$, this region contributes a positive area. The triangle has base $4$ (from $2$ to $6$) and height $8$ (since $f(6) = 2(6) - 4 = 8$):       $\text{Contribution} = \frac{1}{2} \cdot 4 \cdot 8 = 16$

  • Net signed area:     $\int_0^6 (2x - 4)\,dx = -4 + 16 = 12$

• Geometric Computation Example 2: $\int_0^1 (2x - 4)\,dx$:

  • Over the interval $[0, 1]$, the curve lies entirely below the $x$-axis.

  • The geometric shape formed is a trapezoid, broken down into a rectangle and a right triangle:

  • Rectangle over the domain $0 \le x \le 1$ bounded by $-2 \le y \le 0$: area $= 2 \times 1 = 2$.

  • Triangle over the domain $0 \le x \le 1$ bounded between $y = -4$ and $y = -2$: area $= \frac{1}{2} \times 2 \times 1 = 1$.

  • Total geometric magnitude:     $2 \times 1 + \frac{1}{2} \times 2 \times 1 = 3$

  • Because the entire region lies below the $x$-axis, its signed value is negative:     $\int_0^1 (2x - 4)\,dx = -3$

Summation Formulas and Limit Evaluation

• Key Summation Identities:

  • Constant identity:     $\sum_{i=1}^n 1 = n$

  • First powers (integers):     $\sum_{i=1}^n i = \frac{n(n + 1)}{2}$

  • Second powers (squares):     $\sum_{i=1}^n i^2 = \frac{n(n + 1)(2n + 1)}{6}$

  • Third powers (cubes):     $\sum_{i=1}^n i^3 = \frac{n^2(n + 1)^2}{4}$

• Evaluation of $\int_{-1}^1 (x^2 - x)\,dx$ Using the Limit Definition:

  • Interval length: $b - a = 1 - (-1) = 2$.

  • Width of each subinterval for $n$ divisions:     $\Delta x = \frac{2}{n}$

  • Endpoints using a right-hand Riemann sum ($x_i^* = x_i$):     $x_i = a + i\Delta x = -1 + \frac{2i}{n}$

  • Setting up the limit:     $\int_{-1}^1 (x^2 - x)\,dx = \lim_{n \to \infty} \sum_{i=1}^n \left[\left(-1 + \frac{2i}{n}\right)^2 - \left(-1 + \frac{2i}{n}\right)\right]\frac{2}{n}$

  • Expanding the terms inside the summation:     $\left(-1 + \frac{2i}{n}\right)^2 = 1 - \frac{4i}{n} + \frac{4i^2}{n^2}$     $-\left(-1 + \frac{2i}{n}\right) = 1 - \frac{2i}{n}$     $\text{Combined summand} = 2 - \frac{6i}{n} + \frac{4i^2}{n^2}$

  • Distributing $\Delta x = \frac{2}{n}$ across the terms:     $\lim_{n \to \infty} \frac{2}{n} \sum_{i=1}^n \left(2 - \frac{6i}{n} + \frac{4i^2}{n^2}\right) = \lim_{n \to \infty} \left[\frac{4}{n}\sum_{i=1}^n 1 - \frac{12}{n^2}\sum_{i=1}^n i + \frac{8}{n^3}\sum_{i=1}^n i^2\right]$

  • Substituting known summation formulas:     $\lim_{n \to \infty} \left[\frac{4}{n}(n) - \frac{12n(n + 1)}{2n^2} + \frac{8n(n + 1)(2n + 1)}{6n^3}\right]$

  • Expanding the algebraic terms prior to evaluating limits:     $= 4 + \lim_{n \to \infty} \left[-6\left(\frac{n^2 + n}{n^2}\right) + \frac{8}{6}\left(\frac{2n^3 + 3n^2 + n}{n^3}\right)\right]$     $= 4 + \lim_{n \to \infty} \left[-6 - \frac{6}{n} + \frac{16 + \frac{24}{n} + \frac{8}{n^2}}{6}\right]$

  • Evaluating the limit as $n \to \infty$:     $= 4 - 6 + \frac{16}{6}$     $= 4 - 6 + \frac{8}{3}$     $= -2 + \frac{8}{3} = \frac{2}{3}$

Fundamental Properties of the Definite Integral

• Order of Integration: Reversing the bounds of integration negates the value of the integral:   $\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$

• Zero Width Interval: If upper and lower limits of integration are identical, the enclosed area is zero:   $\int_a^a f(x)\,dx = 0$

• Constant Multiple (Scaling): A constant factor $k$ can be factored outside the integral:   $\int_a^b k f(x)\,dx = k\int_a^b f(x)\,dx$

• Sum and Difference: The integral of a sum or difference equals the sum or difference of the individual integrals:   $\int_a^b (f(x) \pm g(x))\,dx = \int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx$

• Additivity of Intervals: Integration over contiguous intervals can be combined into a single integral:   $\int_a^b f(x)\,dx + \int_b^c f(x)\,dx = \int_a^c f(x)\,dx$