Introduction to Definite Integrals and Riemann Sums

Geometric Interpretation of Motion: Velocity and Distance

• Physical Scenario of Motion:

  • A student rides a bicycle toward school, with motion graphed as velocity over time.
  • Time Coordinates: Time is measured along the horizontal axis in units of $\text{minutes}$.
  • Velocity Coordinates: Velocity is measured along the vertical axis in units of $\text{miles per minute}$.
  • Narrative of Motion Phases:
  • Phase 1 (Acceleration): The student starts from rest and accelerates linearly for $5\,\text{minutes}$ until reaching an operating speed of $\frac{1}{6}\,\text{miles per minute}$.
  • Phase 2 (Constant Velocity): The student maintains a steady cruising velocity of $\frac{1}{6}\,\text{miles per minute}$ for an interval of $5\,\text{minutes}$.
  • Phase 3 (Stationary/Stopped): The student comes to a complete halt ($\text{velocity} = 0$) and remains stationary for approximately $5\,\text{minutes}$ after realizing her homework was left at home.
  • Phase 4 (Return Trip - Backward Motion): The student reverses direction (velocity becomes negative relative to the original vector) and travels significantly faster toward home.
  • Phase 5 (Home Search): The student halts at home for approximately $1\frac{2}{3}\,\text{minutes}$ to search for and locate the forgotten homework.
  • Phase 6 (Second Outward Trip): The student accelerates rapidly in the positive direction toward school and sustains cruising speed over a longer interval.

• Relationship Between Speed, Time, and Distance:

  • Basic kinematic distance equation for constant speed: $\text{Distance} = \text{Speed} \times \text{Time}$.
  • For varying velocity curves, total distance traveled corresponds to the geometric area underneath the velocity-time graph.

• Quantitative Evaluation of Traveled Distance:

  • First Triangular Region (Acceleration):
  • $\text{Shape} = \text{Triangle}$
  • $\text{Base} = 5\,\text{minutes}$
  • $\text{Height} = \frac{1}{6}\,\text{miles per minute}$
  • $\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}$
  • $\text{Area}_1 = \frac{1}{2} \times 5 \times \frac{1}{6} = \frac{5}{12}\,\text{miles}$
  • Second Rectangular Region (Steady Riding):
  • $\text{Shape} = \text{Rectangle}$
  • $\text{Base} = 5\,\text{minutes}$
  • $\text{Height} = \frac{1}{6}\,\text{miles per minute}$
  • $\text{Area} = \text{Base} \times \text{Height}$
  • $\text{Area}_2 = 5 \times \frac{1}{6} = \frac{5}{6}\,\text{miles} = \frac{10}{12}\,\text{miles}$
  • Cumulative Outward Distance Traveled:
  • $\text{Total Distance} = \text{Area}_1 + \text{Area}_2 = \frac{5}{12} + \frac{10}{12} = \frac{15}{12}\,\text{miles} = 1.25\,\text{miles}$
  • Return Journey Verification:
  • The backward journey is represented by a triangular area under the zero-axis equal in magnitude to $\frac{15}{12}\,\text{miles}$, confirming that the student returned fully to the starting point (home) rather than recovering the homework along the way.
  • Subsequent Trapezoidal Journey to School:
  • The area of the final trapezoidal region exceeds $\frac{15}{12}\,\text{miles}$, reflecting that the student traversed the entire distance to school.

Generalization to Continuous Curves and Summation Foundations

• Transition from Elementary Geometry to Calculus:

  • Standard geometric formulas readily calculate areas composed of straight-line boundaries (polygons, triangles, rectangles, trapezoids).
  • Arbitrary continuous nonlinear functions lack simple geometric area formulas, requiring systematic accumulation techniques.
  • Parallels across calculus concepts:
  • Rigorous limits require $\epsilon$-$\delta$ formulation.
  • Derivatives require the limit of a difference quotient.
  • Integrals require the infinite limit of accumulated sums.

• Broader Applications of Continuous Integration:

  • Motion tracking: Extracting instantaneous position from velocity, and velocity from acceleration over variable time frames.
  • Probability and Statistics: Discrete distributions rely on standard algebraic summation, whereas continuous probability density functions depend entirely on definite integrals.

• Sum of the First $100$ Positive Integers:

  • Consider the arithmetic series: $1 + 2 + 3 + \dots + 98 + 99 + 100$.
  • Direct manual addition is inefficient; instead, symmetric pairs are formed by combining opposite ends:
  • $1 + 100 = 101$
  • $2 + 99 = 101$
  • $3 + 98 = 101$
  • With $n = 100$, there are $\frac{n}{2} = \frac{100}{2} = 50$ pairs.
  • Total sum evaluation:
  • $\text{Sum} = 101 \times 50 = 5050$
  • Generalized formula for the sum of the first $n$ positive integers:
  • $\text{Sum} = (n + 1) \times \frac{n}{2} = \frac{n(n + 1)}{2}$

• Formal Sigma ($\Sigma$) Notation:

  • Sigma notation condenses extensive or infinite additions into compact algebraic representations:
  • $\sum_{k=m}^n f(k)$
  • Index of Summation: $k$ denotes the running index variable.
  • Lower Limit: $m$ specifies the initial integer step.
  • Upper Limit: $n$ designates the final integer step.
  • Formula/Term: $f(k)$ provides the value generated at each individual index step.
  • Illustrative Expansion Example:
  • Evaluation of $\sum_{k=1}^3 (k^2 + 2)$
  • Step $k = 1$: $1^2 + 2 = 1 + 2 = 3$
  • Step $k = 2$: $2^2 + 2 = 4 + 2 = 6$
  • Step $k = 3$: $3^2 + 2 = 9 + 2 = 11$
  • Sum of steps: $3 + 6 + 11 = 20$

Standard Summation Formulas

• Sum of the First $n$ Natural Numbers:

  • $\sum_{k=1}^n k = \frac{n(n + 1)}{2}$

• Sum of the First $n$ Squares:

  • Expresses the sum of consecutive squared integers ($1 + 4 + 9 + 16 + \dots + n^2$):
  • $\sum_{k=1}^n k^2 = \frac{n(n + 1)(2n + 1)}{6}$

• Sum of the First $n$ Cubes:

  • Expresses the sum of consecutive cubed integers ($1 + 8 + 27 + 64 + \dots + n^3$):
  • $\sum_{k=1}^n k^3 = \frac{n^2(n + 1)^2}{4} = \left( \frac{n(n + 1)}{2} \right)^2$

• Sum of a Constant ($1$):

  • Repeatedly adding the value $1$ across $n$ terms ($1 + 1 + 1 + \dots + 1$):
  • $\sum_{k=1}^n 1 = n$
  • Concrete example for $n = 5$:
  • $\sum_{k=1}^5 1 = 1 + 1 + 1 + 1 + 1 = 5$

Algebraic Properties of Summation and Index Shifting

• Linearity Property (Sum and Difference of Functions):

  • A summation containing a linear combination of functions can be separated into individual summations:
  • $\sum_{k=1}^n [f(k) \pm g(k)] = \sum_{k=1}^n f(k) \pm \sum_{k=1}^n g(k)$

• Constant Multiple Factorization Rule:

  • Any multiplicative constant factor $c$ independent of the index variable $k$ can be factored outside the summation symbol:
  • $\sum_{k=1}^n c \cdot f(k) = c \sum_{k=1}^n f(k)$

• Interval Domain Splitting Rule:

  • For any integer partition point $p$ situated within the interval ($1 \le p \le n$), the summation range can be partitioned into two sub-ranges:
  • $\sum_{k=1}^n f(k) = \sum_{k=1}^{p - 1} f(k) + \sum_{k=p}^n f(k)$
  • Example with $n = 4$ and $p = 3$:
  • $\sum_{k=1}^4 k = \sum_{k=1}^2 k + \sum_{k=3}^4 k = (1 + 2) + (3 + 4) = 3 + 7 = 10$

• Index Transformation (Horizontal Translation):

  • Standard formulas strictly require an initial index of $k = 1$. Summations starting at $k = m > 1$ must be shifted by a change of variables.
  • Shift Rule: To shift the lower bound down by $s$ units to $1$, replace every occurrence of the index $k$ inside the summand with $(k + s)$, and decrease the upper bound by $s$.
  • Application: Transforming $\sum_{k=5}^{1000} (k - 3)$
  • Lower limit needs to shift from $5$ to $1$, requiring a reduction of $s = 4$ units ($5 - 4 = 1$).
  • Upper limit shifts identically: $1000 - 4 = 996$.
  • Substituting $k \rightarrow k + 4$ into the expression:
    • $(k + 4) - 3 = k + 1$
  • Verifying initial terms:
    • Original first term at $k = 5$: $5 - 3 = 2$
    • Shifted first term at $k = 1$: $1 + 1 = 2$
  • Equivalence established: $\sum_{k=5}^{1000} (k - 3) = \sum_{k=1}^{996} (k + 1)$

Step-by-Step Summation Examples and Calculations

• Complete Evaluation of the Shifted Sum:

  • Expression: $\sum_{k=1}^{996} (k + 1)$
  • Splitting via linearity:
  • $\sum_{k=1}^{996} (k + 1) = \sum_{k=1}^{996} k + \sum_{k=1}^{996} 1$
  • Applying known closed-form formulas:
  • $\sum_{k=1}^{996} k = \frac{996 \times (996 + 1)}{2} = \frac{996 \times 997}{2} = 498 \times 997 = 496506$
  • $\sum_{k=1}^{996} 1 = 996$
  • Summing the components:
  • $\text{Total} = 496506 + 996 = 497502$

• Evaluation of a Quadratic Polynomial Summation:

  • Target expression: $\sum_{k=1}^{100} (k^2 + 5k + 6)$
  • Decomposition using summation properties:
  • $\sum_{k=1}^{100} (k^2 + 5k + 6) = \sum_{k=1}^{100} k^2 + 5 \sum_{k=1}^{100} k + 6 \sum_{k=1}^{100} 1$
  • Term 1 ($\sum_{k=1}^{100} k^2$):
  • $\sum_{k=1}^{100} k^2 = \frac{100(100 + 1)(2(100) + 1)}{6} = \frac{100 \times 101 \times 201}{6}$
  • $\frac{2030100}{6} = 338350$
  • Term 2 ($5 \sum_{k=1}^{100} k$):
  • $5 \times \frac{100(101)}{2} = 5 \times 5050 = 25250$
  • Term 3 ($6 \sum_{k=1}^{100} 1$):
  • $6 \times 100 = 600$
  • Cumulative Sum:
  • $\text{Total} = 338350 + 25250 + 600 = 364200$

Riemann Sums and Definite Integral Formulation

• Geometric Area Approximation Strategy:

  • Continuous curves enclose non-polygonal areas that cannot be computed directly using basic Euclidean geometry.
  • Rectangular strips of uniform base width are constructed underneath the curve to approximate total accumulated area.
  • Adding more rectangles of narrower width reduces overestimation and underestimation errors.
  • Taking the limit as the number of subintervals approaches infinity ($n \rightarrow \infty$) yields the exact analytical area.

• Historical Context: Bernhard Riemann:

  • Bernhard Riemann formalized the mathematical framework of accumulating rectangular approximations, known as Riemann sums.
  • Riemann Hypothesis: An unsolved conjecture formulated by Riemann regarding the distribution of prime numbers, remaining an open problem for approximately $160\,\text{years}$ with an associated Millennium Prize of $1{,}000{,}000$.

• General Riemann Sum Definition:

  • $\text{Area} \approx \sum_{k=1}^n f(x_k^*) \Delta x$
  • $\Delta x$ represents the width of each individual rectangular subinterval.
  • $f(x_k^*)$ represents the evaluated height of each rectangle at a selected sample point $x_k^*$ in the $k$-th interval.

• Definition of the Definite Integral:

  • The definite integral is defined formally as the limit of the Riemann sum as the partition count $n$ approaches infinity:
  • $\int_a^b f(x)\,dx = \lim_{n \rightarrow \infty} \sum_{k=1}^n f(x_k^*) \Delta x$

Subinterval Width and Sample Point Selection Strategies

• Uniform Subinterval Width Calculation:

  • Given an overall integration interval $[a, b]$ partitioned into $n$ equal subintervals:
  • $\Delta x = \frac{b - a}{n}$
  • $b$ represents the upper limit endpoint.
  • $a$ represents the lower limit starting point.
  • $n$ represents the total number of approximating rectangles.

• Subinterval Partitions:

  • The interval is partitioned into grid points:
  • $x_0 = a$
  • $x_1 = a + \Delta x$
  • $x_2 = a + 2\Delta x$
  • $x_k = a + k\Delta x$
  • $x_n = b$
  • The $k$-th subinterval is bounded by $[x_{k-1}, x_k]$.

• Strategies for Choosing Sample Point $x_k^*$:

  • Left Endpoint Rule ($L_n$):
  • The evaluation height is fixed at the left boundary of each subinterval.
  • $x_k^* = x_{k-1} = a + (k - 1)\Delta x$
  • Right Endpoint Rule ($R_n$):
  • The evaluation height is fixed at the right boundary of each subinterval.
  • $x_k^* = x_k = a + k\Delta x$
  • Midpoint Rule ($M_n$):
  • The evaluation height is fixed halfway between the boundaries of each subinterval.
  • $x_k^* = \frac{x_{k-1} + x_k}{2} = a + \left( k - \frac{1}{2} \right)\Delta x = a + \frac{2k - 1}{2}\Delta x$
  • Equivalence in the Limit:
  • For any finite $n$, left, right, and midpoint approximations produce distinct numerical totals.
  • As $n \rightarrow \infty$, the subinterval width $\Delta x \rightarrow 0$, collapsing the differences and causing all three sampling techniques to converge to the exact same definite integral value.

Worked Example: Approximating Area Under a Parabolic Curve

• Target Problem Formulation:

  • Function: $f(x) = -x^2 + 4$
  • Domain Interval: Determined by the roots of the function, $[-2, 2]$.
  • Roots determined by Setting $f(x) = 0$:
    • $-x^2 + 4 = 0 \implies x^2 = 4 \implies x = \pm 2$
  • Number of Rectangular Subintervals: $n = 6$

• Step 1: Compute Subinterval Width ($\Delta x$):

  • $a = -2$, $b = 2$, $n = 6$
  • $\Delta x = \frac{b - a}{n} = \frac{2 - (-2)}{6} = \frac{4}{6} = \frac{2}{3}$

• Step 2: Formulate Left Endpoint Sampling ($x_k^*$):

  • General left-hand sample point: $x_k^* = a + (k - 1)\Delta x$
  • Substituting parameters: $x_k^* = -2 + \frac{2}{3}(k - 1)$

• Step 3: Construct Complete Left Riemann Sum Expression:

  • $\text{Area} \approx \sum_{k=1}^6 f\left( -2 + \frac{2}{3}(k - 1) \right) \times \frac{2}{3}$
  • Function substitution into $f(x) = -x^2 + 4$:
  • $\text{Area} \approx \sum_{k=1}^6 \left( -\left( -2 + \frac{2}{3}(k - 1) \right)^2 + 4 \right) \left( \frac{2}{3} \right)$

• Analysis of First Left Subinterval ($k = 1$):

  • At $k = 1$: $x_1^* = -2 + \frac{2}{3}(1 - 1) = -2$
  • Height evaluation: $f(-2) = -(-2)^2 + 4 = -4 + 4 = 0$
  • The first approximating rectangle has an effective height of $0$, appearing as a flat segment along the horizontal axis, which remains geometrically valid in Riemann summation.

Interactive Class Discussions and Dialogue

• Student Observations of the Velocity Graph:

  • Observation on Motion Reversal: Olivia identified that the student riding the bicycle had to travel backwards at a certain stage during the trip.
  • Observation on Speed Changes: A student observed intervals where the cyclist was actively speeding up, as well as intervals spent riding at constant speed.
  • Concept of Area Under Curve: Blake identified that calculating the area under the curve is the mathematical method used to deduce total distance traveled from a velocity-time plot.

• Discussion on Integer Summation Methods:

  • Pairing Strategy: Alana proposed grouping numbers by pairing values like $1 + 99 = 100$ and $2 + 98 = 100$. This concept was refined into end-to-end matching ($1 + 100 = 101$, $2 + 99 = 101$) to account for all $100$ terms uniformly in $50$ identical pairs.
  • Formula Identification: A student recalled the general closed-form equation as $\frac{n(n + 1)}{2}$.

• Summation Notation Clarifications:

  • Index Mechanics Question: A student requested clarification on the role of the variable $k$ in summation expressions.
  • Explanation Provided: The index variable $k$ serves as a step-tracker specifying the starting integer at the bottom of the sigma symbol, incrementing by $1$ through each iteration until terminating at the upper integer limit.
  • Efficiency Question: A student questioned how summation notation improves computational efficiency compared to expanding every term manually.
  • Explanation Provided: While expansions with small upper bounds (such as $n = 3$) seem simple by hand, manual writing becomes impossible when upper bounds extend to $n = 1000$ or $\infty$. Standard summation formulas reduce hundreds or thousands of individual operations into a single algebraic calculation.

• Area Approximation Inquiries:

  • Rectangular Sums: Alana suggested filling the irregular region beneath a curve with small rectangles and summing their areas to estimate total area.
  • Integral Notation: Olivia described the definite integral symbol as an elongated "$\text{S}$" with the upper root at the top and the lower root at the bottom.
  • Software Graphing Oddity: A numerical artifact in Desmos produced an erroneous negative display during area accumulation approximations, which was noted as an internal computational quirk, as true area approximations under positive functions must approach positive values near $10$.

• Administrative Class Policies:

  • Homework assignments are published via Blackboard.
  • Deadlines are subject to extensions depending on instructional pacing and whether all planned lecture topics are fully addressed during the instructional week.