MATH1045 Calculus Study Guide: Rates of Change, Definite Integrals, and Accumulation Functions

Analysis of Rates of Change and Derivations of Net Change

• The fundamental relationship between a quantity function $f(t)$, its rate of change (first derivative) $f'(t)$, and its second derivative $f''(t)$ determines the behavior of a physical, economic, or biological system:

  • When the original function $f(t)$ is positive ($f(t) > 0$), the state variable has a positive magnitude above the reference axis.
  • When the first derivative $f'(t)$ is positive ($f'(t) > 0$), the original function $f(t)$ is increasing.
  • When the first derivative $f'(t)$ is negative ($f'(t) < 0$), the original function $f(t)$ is decreasing.
  • When the first derivative $f'(t)$ is zero ($f'(t) = 0$), the original function $f(t)$ is flat (stationary point).
  • When the second derivative $f''(t)$ is positive ($f''(t) > 0$), the first derivative $f'(t)$ is increasing, meaning the original function $f(t)$ is concave up.
  • When the second derivative $f''(t)$ is negative ($f''(t) < 0$), the first derivative $f'(t)$ is decreasing, meaning the original function $f(t)$ is concave down.

• Evaluating Function Behavior at Specific Points:

  • Given a function $f(t)$ with labeled points $A$ and $B$:
  • At Point $A$ (where the curve lies below the horizontal axis, slopes upward, and is concave up):
    • Original function $f(t)$: negative
    • First derivative $f'(t)$: positive
    • Original function $f(t)$ direction: increasing
    • First derivative $f'(t)$ direction: increasing
  • At Point $B$ (where the curve lies above the horizontal axis at a local peak):
    • Original function $f(t)$: positive
    • First derivative $f'(t)$: zero
    • Second derivative $f''(t)$: negative
    • Original function $f(t)$ direction: flat
    • First derivative $f'(t)$ direction: decreasing

• Analyzing Function Behavior from Derivative Graphs:

  • Given a graph of a derivative $f'(x)$, specific features translate directly to the original function $f(x)$:
  • Points where $f'(x)$ crosses from negative to positive correspond to relative minima of $f(x)$.
  • Points where $f'(x)$ crosses from positive to negative correspond to relative maxima of $f(x)$.
  • Local extrema (peaks and troughs) of $f'(x)$ correspond to inflection points of $f(x)$, where concavity changes.
  • At Point $A$ on a derivative curve $f'(x)$ located below the horizontal axis and sloping upward:
    • First derivative $f'(x)$: negative
    • Second derivative $f''(x)$: positive
    • Original function $f(x)$: decreasing
    • First derivative $f'(x)$: increasing
  • At Point $B$ on a derivative curve $f'(x)$ located at a positive local maximum:
    • First derivative $f'(x)$: positive
    • Second derivative $f''(x)$: zero
    • Original function $f(x)$: increasing
    • First derivative $f'(x)$: flat

• Antiderivative Relationships:

  • An antiderivative $F(x)$ of a continuous function $f(x)$ satisfies the differential relation:     $F'(x) = f(x)$
  • The indefinite integral represents the family of all antiderivatives:     $\int f(x)\,dx = F(x) + C$

Riemann Sums and Finite Rectangle Approximations

• Fundamental Definitions in Rectangle Approximations:

  • When approximating the signed area between the graph of a rate function $f'(x)$ and the $x$-axis over an interval $[a, b]$:
  • $n$ represents the total number of subintervals (or approximating rectangles).
  • $\Delta x$ represents the width (grid spacing) of each uniform subinterval:       $\Delta x = \frac{b - a}{n}$

• Fundamental Principle of Net Change via Area:

  • The signed area bounded by the derivative graph $f'(x)$ from $x = a$ to $x = b$ equals the total net accumulated change in the original function $f(x)$:     $\int_{a}^{b} f'(x)\,dx = f(b) - f(a)$

• Accuracy of Riemann Sum Approximations:

  • When using a Riemann sum with $n$ rectangles to approximate a definite integral $\int_{a}^{b} f(x)\,dx$:
  • Decreasing the subinterval width $\Delta x$ yields a better, more precise approximation.
  • Increasing the number of rectangles $n$ yields a better, more precise approximation.
  • For instance, approximating $\int_{0}^{5} f(x)\,dx$ with $\Delta x = 0.05$ provides a far more accurate result than $\Delta x = 0.5$, $\Delta x = 5$, $\Delta x = 50$, or $\Delta x = 500$.
  • Similarly, choosing $n = 500$ rectangles provides a far superior approximation compared to $n = 50$, $n = 5$, $n = 0.5$, or $n = 0.05$.

• Calculating Grid Parameters:

  • For an approximation of accumulated change on $[-1, 2]$ using right rectangles with $\Delta x = 0.5$:     $n = \frac{b - a}{\Delta x} = \frac{2 - (-1)}{0.5} = \frac{3}{0.5} = 6$
  • For an approximation of accumulated change on $[-2, 2.5]$ using left rectangles with $n = 3$:     $\Delta x = \frac{b - a}{n} = \frac{2.5 - (-2)}{3} = \frac{4.5}{3} = 1.5$

• Direction and Over/Under-Estimation Rules:

  • For a strictly decreasing function over an interval:
  • Left Riemann sums yield an overestimate of the true integral because the top border of each rectangle lies above the curve.
  • Right Riemann sums yield an underestimate of the true integral because the top border of each rectangle lies below the curve.
  • For a strictly increasing function over an interval:
  • Left Riemann sums yield an underestimate.
  • Right Riemann sums yield an overestimate.

• Analysis of Power Consumption Approximation:

  • Given a graph showing rectangles constructed under a power consumption rate curve $g(t)$ (measured in kilowatts per hour or kW):
  • Subinterval width: $\Delta t = 1\,\text{hour}$
  • Total subintervals shown across $t = 0$ to $t = 12$: $n = 12$
  • Area units: $\text{kilowatts per hour} \times \text{hours} = \text{kilowatts}$
  • Type of sum: Right rectangle approximation (heights defined by right endpoints of subintervals).
  • Estimation: Under-estimates the definite integral because the curve is generally decreasing over the domain.
  • Physical Meaning: Approximates the net change in total power consumption over the time interval.

• Tabular Riemann Sum Computations:

  • Given discrete data for rowboat leak rate $r(t)$ in gallons per minute:

• Parameters:

  • Number of subintervals: $n = 4$
  • Subinterval width: $\Delta t = 5\,\text{min}$

• Left Riemann Sum Calculation:     $\text{Left Sum} = \Delta t \times [r(0) + r(5) + r(10) + r(15)] = 5 \times (4 + 9 + 12 + 13) = 5 \times 38 = 190\,\text{gallons}$

• Right Riemann Sum Calculation:     $\text{Right Sum} = \Delta t \times [r(5) + r(10) + r(15) + r(20)] = 5 \times (9 + 12 + 13 + 10) = 5 \times 44 = 220\,\text{gallons}$

  • Discrete Index Summation Example:

• Given $f(t) = t^2 - 3t$ and sample points:

  • $t_1 = -2$
  • $t_2 = 1$
  • $t_3 = 4$
  • $t_4 = 7$
  • $t_5 = 10$

• Step (a): Step size $\Delta t = t_2 - t_1 = 1 - (-2) = 3$

• Step (b): Compute $\sum_{i=2}^{4} f(t_i) \cdot \Delta t$:

  • $f(t_2) = f(1) = 1^2 - 3(1) = -2$
  • $f(t_3) = f(4) = 4^2 - 3(4) = 4$
  • $f(t_4) = f(7) = 7^2 - 3(7) = 28$
  • $\sum_{i=2}^{4} f(t_i) \cdot \Delta t = (-2 + 4 + 28) \times 3 = 30 \times 3 = 90$

Definite Integrals: Exact Calculation, Geometry, and Properties

• Fundamental Integral Properties:
  • Reversing Integration Limits:     $\int_{a}^{b} f(x)\,dx = -\int_{b}^{a} f(x)\,dx$
  • Additivity over Adjacent Intervals:     $\int_{a}^{c} f(x)\,dx = \int_{a}^{b} f(x)\,dx + \int_{b}^{c} f(x)\,dx$

• Geometric Integration of Piecewise Linear Functions:

  • Given a graph $f(x)$ defined on $[-10, 70]$ with vertices at $(-10, 4)$, $(0, 8)$, $(30, -4)$, $(40, -4)$, $(60, 0)$, and $(70, 2)$:
  • Definite Integral $\int_{-10}^{30} f(x)\,dx$:
    • Region 1 ($[-10, 0]$): Trapezoid above axis with parallel vertical heights 4 and 8, width 10:         $\text{Area}_1 = \frac{4 + 8}{2} \times 10 = 60$
    • Region 2 ($[0, 30]$): Line segment from $(0, 8)$ to $(30, -4)$ crosses the $x$--axis at $x = 20$:
    • Triangle above axis ($[0, 20]$): $\frac{1}{2} \times 20 \times 8 = 80$
    • Triangle below axis ($[20, 30]$): $\frac{1}{2} \times 10 \times (-4) = -20$
    • Total Integral: $60 + 80 - 20 = 120$
  • Definite Integral $\int_{30}^{70} f(x)\,dx$:
    • Region 1 ($[30, 40]$): Rectangle below axis of height 4 and width 10:         $\text{Area} = 10 \times (-4) = -40$
    • Region 2 ($[40, 60]$): Triangle below axis of base 20 and height 4:         $\text{Area} = \frac{1}{2} \times 20 \times (-4) = -40$
    • Region 3 ($[60, 70]$): Triangle above axis of base 10 and height 2:         $\text{Area} = \frac{1}{2} \times 10 \times 2 = 10$
    • Total Integral: $-40 + (-40) + 10 = -70$
  • Definite Integral $\int_{-10}^{70} f(x)\,dx$:       $\int_{-10}^{70} f(x)\,dx = \int_{-10}^{30} f(x)\,dx + \int_{30}^{70} f(x)\,dx = 120 + (-70) = 50$

• Additional Piecewise Linear Definite Integral Worked Example:

  • Given a rate function $f'(x)$ with line segments joining $(0, -15)$, $(10, 0)$, $(20, 15)$, $(30, 15)$, $(40, 0)$, and $(50, 15)$:
  • To calculate exact change in $f(x)$ from $x = 0$ to $x = 40$:
  • Interval $[0, 10]$: Triangle below axis: $\frac{1}{2} \times 10 \times (-15) = -75$
  • Interval $[10, 20]$: Triangle above axis: $\frac{1}{2} \times 10 \times 15 = 75$
  • Interval $[20, 30]$: Rectangle above axis: $10 \times 15 = 150$
  • Interval $[30, 40]$: Triangle above axis: $\frac{1}{2} \times 10 \times 15 = 75$
  • Accumulated change $\Delta f = -75 + 75 + 150 + 75 = 225$

• Comparing Definite Integral Values from Enclosed Areas:

  • Given a continuous function $g(x)$ over $[0, 10]$ creating four regions bounded by the axis with absolute areas:
  • Region 1 ($[0, 2]$): Area above axis = 8
  • Region 2 ($[2, 5]$): Area below axis = 11
  • Region 3 ($[5, 8]$): Area above axis = 6
  • Region 4 ($[8, 10]$): Area below axis = 7
  • Integral Calculations:
  • $\int_{0}^{8} g(x)\,dx = (+8) + (-11) + (+6) = 3$
  • $\int_{2}^{10} g(x)\,dx = (-11) + (+6) + (-7) = -12$
  • $\int_{2}^{5} g(x)\,dx = -11$
  • $\int_{8}^{10} g(x)\,dx = -7$
  • $\int_{0}^{10} g(x)\,dx = 8 - 11 + 6 - 7 = -4$
  • Ranking from smallest to largest:
  • $\int_{2}^{10} g(x)\,dx = -12$
  • $\int_{2}^{5} g(x)\,dx = -11$
  • $\int_{8}^{10} g(x)\,dx = -7$
  • $\int_{0}^{10} g(x)\,dx = -4$
  • $\int_{0}^{8} g(x)\,dx = 3$
  • The largest value among these definite integrals is $\int_{0}^{8} g(x)\,dx = 3$

Real-World Applications and Contextual Interpretations

• Environmental and Industrial Rates:

  • Chemical Dumping Interpretation:
  • For dumping rate $P(t)$ in tons per year where $t$ is years since 2009, the statement $\int_{1}^{7} P(t)\,dt = 4250$ means:
  • Between the year 2010 ($t = 1$) and the year 2016 ($t = 7$), the factory dumped a total quantity of 4250 tons of dangerous chemical into the lake.
  • Oil Tanker Leakage:
  • For oil leaking at $g(t)$ barrels per hour, the total quantity of oil leaked during the first full day (24 hours) is expressed as:       $\int_{0}^{24} g(t)\,dt$
  • Using a left sum with $\Delta t = 6$ hours on the data set $g(0)=1.2, g(6)=3.8, g(12)=7.9, g(18)=11.2, g(24)=12.7, g(30)=12.9$ yields:       $\text{Total Oil} \approx 6 \times (1.2 + 3.8 + 7.9 + 11.2) = 6 \times 24.1 = 144.6\,\text{barrels}$

• Biological Growth Models:

  • Fungus Growth Rate Interpretation:
  • If $w(t)$ is the growth rate of a fungus in mg per hour at time $t$ hours:
    • $w(5) = 0.3$: At exactly 5 hours after being placed on the growth medium, the fungus is growing at a rate of $0.3\,\text{mg per hour}$.
    • $\int_{0}^{5} w(t)\,dt = 4$: During the first 5 hours after placement on the growth medium, the total mass of the fungus increased by $4\,\text{mg}$.
  • Bacteria Colony Expansion:
  • An initial population of 2 million bacteria grows at $f(t) = 3e^{0.04t}$ million bacteria per hour.
  • Growth in the first 4 hours:       $\Delta P = \int_{0}^{4} 3e^{0.04t}\,dt = \left[ \frac{3}{0.04} e^{0.04t} \right]_{0}^{4} = 75 \left( e^{0.16} - 1 \right) \approx 13.013\,\text{million bacteria}$
  • Total population after 4 hours:       $P(4) = 2 + 13.013 = 15.013\,\text{million bacteria}$
  • Eucalyptus Tree Growth:
  • Height rate $h(t) = 0.6 + \frac{4}{(t+1)^3}$ feet per year.
  • Height of a 3-year-old tree starting from 0 ft height:       $h_{\text{total}} = \int_{0}^{3} \left( 0.6 + \frac{4}{(t+1)^3} \right)\,dt = \left[ 0.6t - \frac{2}{(t+1)^2} \right]_{0}^{3} = \left( 1.8 - \frac{2}{16} \right) - (0 - 2) = 3.675\,\text{feet}$
  • Definite integral $\int_{4}^{7} h(t)\,dt$:       $\int_{4}^{7} h(t)\,dt = \left[ 0.6t - \frac{2}{(t+1)^2} \right]_{4}^{7} = \left( 4.2 - \frac{2}{64} \right) - \left( 2.4 - \frac{2}{25} \right) = 4.16875 - 2.32 = 1.84875\,\text{feet}$
    • Interpretation: Between year 4 and year 7, the height of the eucalyptus tree increases by 1.84875 feet.

• Kinematic & Distance Applications:

  • Free-Falling Skydiver Distance Model:
  • Velocity $v(t) = 49 - 49(0.8187^t)\,\text{m/s}$.
  • Integral for total distance fallen after $t$ seconds:       $D(t) = \int_{0}^{t} \left( 49 - 49(0.8187^x) \right)\,dx$
  • Total distance to ground from 6000 meters altitude is reached when $D(t) = 6000\,\text{meters}$, which solves to approximately $t \approx 133\,\text{seconds}$.
  • Particle Position Change:
  • Velocity curve $v(t) = 6 - 2t\,\text{cm/sec}$ on a straight path:
    • Position change from $t = 0$ to $t = 3$: Triangle area above axis = $\frac{1}{2} \times 3 \times 6 = 9\,\text{cm}$ to the right.
    • Position change from $t = 0$ to $t = 5$: Triangle above axis ($+9\,\text{cm}$) plus triangle below axis ($\frac{1}{2} \times 2 \times (-4) = -4\,\text{cm}$) gives net position change of $9 - 4 = 5\,\text{cm}$.

• Economic and Financial Models:

  • Toy Price Change:
  • Given price rate $t(x)$ in dollars per month, area units on the rate graph are $\text{dollars per month} \times \text{months} = \text{dollars}$.
  • With region area 10 below axis ($x \in [0, 3]$), region area 13 above axis ($x \in [3, 7]$), and region area 17 below axis ($x \in [7, 10]$):
    • First 3 months: Price decreased by 10 dollars.
    • After 7 months: Net price change is $-10 + 13 = +3$ dollars (price is 3 dollars higher than at $t=0$).
    • After 10 months: Net price change is $-10 + 13 - 17 = -14$ dollars (price is 14 dollars lower than at $t=0$).
  • Cell-Phone Bill Rate Analysis:
  • Rate function $C(t)$ in dollars per year, where $t$ is years since 2000:
    • Axis units: $t$ in years, $C(t)$ in dollars per year.
    • Bounded Area Units: $\text{dollars per year} \times \text{years} = \text{dollars}$.
    • Decreasing bill period: Between years 2004 ($t=4$) and 2010 ($t=10$), where rate $C(t) < 0$
    • Lowest average bill occurrence: At year 2010 ($t=10$).
    • Overall change from 2000 to 2016 ($t=0$ to $t=16$):         $\text{Net Change} = \int_{0}^{16} C(t)\,dt = (+8) + (-8) + (-4) + (+36) = +32\,\text{dollars}$

• Industrial Soda Production Rates:

  • Comparing Soda Companies A and B over a 10-hour window:
  • Company B reaches a higher maximum peak rate than Company A.
  • The overall total production of Company A exceeds Company B because the integral $\int_{0}^{10} A(t)\,dt$ captures greater accumulated area than $\int_{0}^{10} B(t)\,dt$.

Accumulation Functions

• Definition and Basic Properties:
  • An accumulation function $A(x)$ represents the net signed area under a continuous function $f(t)$ from a fixed basepoint $a$ to a variable upper limit $x$:     $A(x) = \int_{a}^{x} f(t)\,dt$
  • Evaluating at the basepoint always yields zero:     $A(a) = \int_{a}^{a} f(t)\,dt = 0$

• Mapping Relationships Between $f(t)$ and $A(x)$:
  • Sign of $f(t)$ determines Monotonicity of $A(x)$:
  • When $f(t) > 0$, $A(x)$ is strictly increasing.
  • When $f(t) < 0$, $A(x)$ is strictly decreasing.
  • When $f(t) = 0$, $A(x)$ has stationary critical points (local extrema or plateau points).
  • Monotonicity of $f(t)$ determines Concavity of $A(x)$:
  • When $f(t)$ is increasing ($f'(t) > 0$), $A(x)$ is concave up.
  • When $f(t)$ is decreasing ($f'(t) < 0$), $A(x)$ is concave down.
  • When $f(t)$ reaches a local extremum ($f'(t) = 0$), $A(x)$ has an inflection point.

• Step-by-Step Evaluation of Accumulation Values:

  • Given $\text{Area}_1 = 10$ below axis on $[0, a]$, $\text{Area}_2 = 13$ above axis on $[a, b]$, and $\text{Area}_3 = 17$ below axis on $[b, c]$:
  • For the accumulation function $A(x) = \int_{0}^{x} f(t)\,dt$:
  • $A(0) = \int_{0}^{0} f(t)\,dt = 0$
  • $A(a) = \int_{0}^{a} f(t)\,dt = -10$
  • $A(b) = \int_{0}^{b} f(t)\,dt = -10 + 13 = 3$
  • $A(c) = \int_{0}^{c} f(t)\,dt = 3 - 17 = -14$

• Complete Analytic Accumulation Investigation:

  • Consider $f(t) = 4 - t$ over the domain $0 \le t \le 6$:
  • Defining the accumulation function:     $A(x) = \int_{0}^{x} (4 - t)\,dt = \left[ 4t - \frac{t^2}{2} \right]_{0}^{x} = 4x - \frac{x^2}{2}$
  • Table of Values for $A(x)$:

• Observations:
  • $A(x)$ increases on $[0, 4]$ where $f(t) > 0$.
  • $A(x)$ reaches its maximum value of 8 at $x = 4$ where $f(4) = 0$.
  • $A(x)$ decreases on $[4, 6]$ where $f(t) < 0$.
  • $A(x)$ is concave down across the entire domain because $f(t) = 4 - t$ is strictly decreasing.