Solved Exercises on Accumulations of Change Introduction

Master accumulations of change: definite integrals, area under curves, Riemann sums, net change theorem, and applications to physics problems through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Area under curve
Exercise 1
Find the area under the curve \(f(x) = 2x + 1\) from \(x = 0\) to \(x = 3\) using the definite integral.
Definition:

Accumulation of change: The definite integral \(\int_a^b f(x)dx\) represents the accumulated change in a quantity whose rate of change is given by \(f(x)\).

Integration method:
  1. Identify the function \(f(x)\) and integration bounds \([a,b]\)
  2. Find the antiderivative \(F(x)\) of \(f(x)\)
  3. Apply the Fundamental Theorem of Calculus: \(\int_a^b f(x)dx = F(b) - F(a)\)
Integral
\(\int_0^3 (2x + 1)dx\)
Antiderivative
\(x^2 + x\)
Result
\(12\)
Step 1: Set up the integral

We want to find the area under \(f(x) = 2x + 1\) from \(x = 0\) to \(x = 3\)

\(\int_0^3 (2x + 1)dx\)

Step 2: Find the antiderivative

\(\int (2x + 1)dx = \int 2x dx + \int 1 dx = x^2 + x + C\)

Step 3: Apply Fundamental Theorem of Calculus

\(\int_0^3 (2x + 1)dx = [x^2 + x]_0^3\)

\(= (3^2 + 3) - (0^2 + 0) = (9 + 3) - 0 = 12\)

\(\int_0^3 (2x + 1)dx = 12\)
Final answer:

The area under the curve from \(x = 0\) to \(x = 3\) is 12 square units.

Applied rules:

• Fundamental Theorem of Calculus: Connects definite integrals to antiderivatives

• Power Rule: \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\) (for \(n \neq -1\))

• Linearity: \(\int [af(x) + bg(x)]dx = a\int f(x)dx + b\int g(x)dx\)

2 Net change in velocity
Exercise 2
A car's acceleration is given by \(a(t) = 4t - 2\) m/s². Find the change in velocity from \(t = 1\) to \(t = 4\) seconds.
Definition:

Net change theorem: The integral of a rate of change gives the net change in the quantity: \(\int_a^b F'(x)dx = F(b) - F(a)\)

Acceleration
\(a(t) = 4t - 2\)
Velocity integral
\(\int_1^4 (4t - 2)dt\)
Change in velocity
\(15\) m/s
Step 1: Understand the relationship

Since acceleration is the derivative of velocity (\(a(t) = v'(t)\)), the change in velocity is the integral of acceleration.

\(\Delta v = \int_1^4 a(t)dt = \int_1^4 (4t - 2)dt\)

Step 2: Find the antiderivative

\(\int (4t - 2)dt = \int 4t dt - \int 2 dt = 2t^2 - 2t + C\)

Step 3: Apply the Net Change Theorem

\(\int_1^4 (4t - 2)dt = [2t^2 - 2t]_1^4\)

\(= (2(4)^2 - 2(4)) - (2(1)^2 - 2(1))\)

\(= (32 - 8) - (2 - 2) = 24 - 0 = 24\)

\(\Delta v = 24\) m/s
Final answer:

The car's velocity increases by 24 m/s from \(t = 1\) to \(t = 4\) seconds.

Applied rules:

• Net Change Theorem: \(\int_a^b F'(x)dx = F(b) - F(a)\)

• Physics Application: Velocity is the integral of acceleration

• Power Rule: \(\int t^n dt = \frac{t^{n+1}}{n+1} + C\)

3 Water accumulation
Exercise 3
Water flows into a tank at a rate of \(r(t) = 3t^2 + 2\) liters per minute. Find the total amount of water added from \(t = 0\) to \(t = 2\) minutes.
Definition:

Accumulated quantity: When a rate of change is given, the definite integral gives the total change over an interval.

Rate function
\(r(t) = 3t^2 + 2\)
Accumulation
\(\int_0^2 (3t^2 + 2)dt\)
Total water
\(8\) liters
Step 1: Set up the accumulation integral

The rate of flow is \(r(t) = 3t^2 + 2\) liters per minute. The total accumulation from \(t = 0\) to \(t = 2\) is:

\(\int_0^2 (3t^2 + 2)dt\)

Step 2: Find the antiderivative

\(\int (3t^2 + 2)dt = \int 3t^2 dt + \int 2 dt = t^3 + 2t + C\)

Step 3: Evaluate the definite integral

\(\int_0^2 (3t^2 + 2)dt = [t^3 + 2t]_0^2\)

\(= (2^3 + 2(2)) - (0^3 + 2(0)) = (8 + 4) - 0 = 12\)

\(\int_0^2 (3t^2 + 2)dt = 12\) liters
Final answer:

12 liters of water are added to the tank from \(t = 0\) to \(t = 2\) minutes.

Applied rules:

• Accumulation Principle: Rate × Time = Total Amount

• Integration of polynomials: Use power rule for each term

• Definite integral evaluation: Apply FTC

Rules and methods, laws,...
\(\int_a^b f(x)dx = F(b) - F(a)\)
Fundamental Theorem of Calculus
Net Change Theorem
\(\int_a^b F'(x)dx = F(b) - F(a)\)
Integral of rate of change equals net change
Power Rule
\(\int x^n dx = \frac{x^{n+1}}{n+1} + C\)
Integration of powers (n ≠ -1)
Linearity
\(\int [af(x) + bg(x)]dx = a\int f(x)dx + b\int g(x)dx\)
Integration is linear operation
Key definitions:

Accumulation of change: The process of adding up small changes over an interval to find the total change

Definite integral: Represents the signed area under a curve and the accumulated change

Rate of change: Function describing how quickly a quantity changes (derivative)

Complete methodology:
  1. Identify the rate function: Determine what rate of change is given (velocity, acceleration, flow rate, etc.)
  2. Set up the integral: Write the definite integral with appropriate bounds
  3. Find the antiderivative: Use integration rules to find F(x)
  4. Evaluate: Apply the Fundamental Theorem of Calculus
Tip 1: Always identify what the rate function represents (velocity, acceleration, flow rate, etc.)
Tip 2: The integral of velocity gives displacement, the integral of acceleration gives velocity.
Tip 3: Check units: if rate is in meters/second, integral gives meters.
Tip 4: Draw a sketch of the function to visualize the area being calculated.
Common errors: Forgetting to apply FTC correctly, misidentifying the rate function, incorrect antiderivative calculation.
Physical interpretation: Definite integrals represent accumulated quantities in physical contexts.
Solution: Exercises 4 to 5
4 Area between curves
Exercise 4
Find the area between the curves \(f(x) = x^2\) and \(g(x) = x\) from \(x = 0\) to \(x = 1\).
Definition:

Area between curves: \(\int_a^b |f(x) - g(x)|dx\) where the absolute value ensures positive area.

Function difference
\(x - x^2\)
Integral setup
\(\int_0^1 (x - x^2)dx\)
Area
\(\frac{1}{6}\)
Step 1: Determine which function is greater

On the interval [0,1], we need to check which function is larger. Testing \(x = 0.5\):

\(f(0.5) = (0.5)^2 = 0.25\) and \(g(0.5) = 0.5\)

So \(g(x) > f(x)\) on [0,1], meaning the area is \(\int_0^1 (g(x) - f(x))dx = \int_0^1 (x - x^2)dx\)

Step 2: Find the antiderivative

\(\int (x - x^2)dx = \int x dx - \int x^2 dx = \frac{x^2}{2} - \frac{x^3}{3} + C\)

Step 3: Evaluate the definite integral

\(\int_0^1 (x - x^2)dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1\)

\(= \left(\frac{1}{2} - \frac{1}{3}\right) - (0 - 0) = \frac{3-2}{6} = \frac{1}{6}\)

Area = \(\frac{1}{6}\) square units
Final answer:

The area between the curves \(y = x^2\) and \(y = x\) from \(x = 0\) to \(x = 1\) is \(\frac{1}{6}\) square units.

Applied rules:

• Area between curves: \(\int_a^b |f(x) - g(x)|dx\)

• Comparison test: Determine which function is greater on the interval

• Integration of polynomials: Use power rule for each term

5 Temperature change
Exercise 5
The rate of temperature change in a chemical reaction is given by \(r(t) = 6e^{-0.5t}\) degrees Celsius per minute. Find the total temperature change from \(t = 0\) to \(t = 4\) minutes.
Definition:

Exponential decay rate: Functions of the form \(ae^{kt}\) where \(k < 0\) model decaying rates.

Rate function
\(6e^{-0.5t}\)
Antiderivative
\(-12e^{-0.5t}\)
Temperature change
\(10.37\)°C
Step 1: Set up the integral

The rate of temperature change is \(r(t) = 6e^{-0.5t}\). The total change from \(t = 0\) to \(t = 4\) is:

\(\int_0^4 6e^{-0.5t}dt\)

Step 2: Find the antiderivative

For \(\int 6e^{-0.5t}dt\), use the formula \(\int e^{kt}dt = \frac{1}{k}e^{kt} + C\)

\(\int 6e^{-0.5t}dt = 6 \cdot \frac{1}{-0.5}e^{-0.5t} = -12e^{-0.5t} + C\)

Step 3: Evaluate the definite integral

\(\int_0^4 6e^{-0.5t}dt = [-12e^{-0.5t}]_0^4\)

\(= -12e^{-0.5(4)} - (-12e^{-0.5(0)})\)

\(= -12e^{-2} + 12e^0 = -12e^{-2} + 12\)

\(= 12(1 - e^{-2}) = 12(1 - \frac{1}{e^2}) \approx 12(1 - 0.135) \approx 10.37\)

Step 4: Interpret the result

The temperature increases by approximately 10.37°C during the reaction.

Total temperature change ≈ 10.37°C
Final answer:

The total temperature change from \(t = 0\) to \(t = 4\) minutes is approximately 10.37°C.

Applied rules:

• Exponential integration: \(\int e^{kt}dt = \frac{1}{k}e^{kt} + C\)

• Constant multiple: \(\int af(x)dx = a\int f(x)dx\)

• Numerical approximation: \(e^{-2} \approx 0.135\)

Key Concepts, Laws, Methods, and Formulas
\(\int_a^b f(x)dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x\)
Definition of Definite Integral
Key definitions:

Accumulation function: \(A(x) = \int_a^x f(t)dt\) represents the accumulated change from \(a\) to \(x\)

Net change: The difference between final and initial values of a quantity

Rate of change: The derivative of a function that describes how a quantity changes

Complete methodology:
  1. Problem identification: Recognize when a rate of change is given and total change is needed
  2. Setup: Write the definite integral with correct limits
  3. Integration: Find the antiderivative using appropriate rules
  4. Evaluation: Apply FTC to get the numerical result
  5. Interpretation: Relate the result back to the original context
Tip 1: The definite integral \(\int_a^b f(x)dx\) gives the net accumulation, not just the area.
Tip 2: Areas below the x-axis contribute negatively to the total accumulation.
Tip 3: In physics, position is the integral of velocity, velocity is the integral of acceleration.
Tip 4: Always consider the units when interpreting results.
Applications: Physics (motion), economics (cost/revenue), biology (population growth), chemistry (reaction rates).
Integration rules: Power rule, exponential rule, logarithmic rule, trigonometric integrals.
Formulas to know by heart:

• Fundamental Theorem: \(\int_a^b f(x)dx = F(b) - F(a)\)

• Power rule: \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\) (for \(n \neq -1\))

• Exponential: \(\int e^{kx} dx = \frac{1}{k}e^{kx} + C\)

• Constant: \(\int k dx = kx + C\)

• Net change: \(\int_a^b F'(x)dx = F(b) - F(a)\)

Exercise with Visualization: Accumulation Functions
Exercise 6: Accumulation Analysis
Consider the rate functions:
\(r_1(t) = 2t\) (linear growth)
\(r_2(t) = 3e^{0.2t}\) (exponential growth)
\(r_3(t) = 5\cos(t)\) (oscillating)

Analysis: The chart shows how different rate functions lead to different accumulation patterns.

  • \(r_1(t) = 2t\) → Accumulation: \(t^2\) (quadratic growth)
  • \(r_2(t) = 3e^{0.2t}\) → Accumulation: \(15(e^{0.2t} - 1)\) (exponential growth)
  • \(r_3(t) = 5\cos(t)\) → Accumulation: \(5\sin(t)\) (oscillating pattern)

Questions & Answers

Question: I'm confused about when to use the Net Change Theorem versus just evaluating a definite integral. Are they the same thing?

Answer: Great question! They are closely related but emphasize different aspects:

  • The Fundamental Theorem of Calculus states that \(\int_a^b f(x)dx = F(b) - F(a)\) where \(F\) is an antiderivative of \(f\).
  • The Net Change Theorem is a specific application of FTC that emphasizes the physical meaning: \(\int_a^b F'(x)dx = F(b) - F(a)\), which represents the net change in \(F\) from \(a\) to \(b\).

In essence, the Net Change Theorem is FTC applied specifically to a situation where you know the rate of change (\(F'\)) and want the total change (\(F(b) - F(a)\)).

Example: If \(v(t)\) is velocity (rate of change of position), then \(\int_a^b v(t)dt = s(b) - s(a)\) is the change in position (displacement).

Question: How do I know if I should take the absolute value when calculating areas between curves?

Answer: The choice depends on whether you want signed area or total area:

  • For total geometric area (always positive), use: \(\int_a^b |f(x) - g(x)|dx\)
  • For net area (accounting for regions above and below the x-axis), use: \(\int_a^b [f(x) - g(x)]dx\)

In practice, you usually determine which function is greater on the interval by testing a point, then compute \(\int_a^b (\text{top function} - \text{bottom function})dx\).

If the curves cross within the interval, you'll need to break the integral at intersection points and sum the absolute values of each segment.

Tip: Sketch the functions first to visualize which is on top!

Question: I often get confused about units when doing accumulation problems. How can I keep track of them?

Answer: Unit tracking is crucial! Here's a systematic approach:

  • If the rate function has units of "Y per X", then the integral has units of "Y"
  • For example: velocity (m/s) integrated over time (s) gives displacement (m)
  • Flow rate (L/min) integrated over time (min) gives volume (L)

Always write out units during calculations:

\(\int_0^5 \text{(velocity in m/s)} \, dt \text{ (in s)} = \text{displacement in m}\)

This dimensional analysis helps catch errors and confirms your answer makes physical sense!

Question: How do I handle negative values in accumulation problems? Do they mean something went wrong?

Answer: Negative values in accumulation problems are completely normal and meaningful!

They indicate:

  • Negative rate: The quantity is decreasing (e.g., cooling, draining)
  • Direction: In physics, negative displacement means opposite direction
  • Reduction: The total amount decreases over that interval

For example, if velocity is negative, the object moves backward. If a rate of water flow is negative, water is leaving the system.

The definite integral gives the net change, which can be positive, negative, or zero depending on the balance of positive and negative contributions.

To find total change regardless of direction, integrate the absolute value of the rate function.

Question: What's the difference between average rate and accumulated change, and how do they relate to integrals?

Answer: These are related but distinct concepts:

  • Accumulated change: \(\int_a^b f(x)dx\) - total change over the interval
  • Average rate: \(\frac{1}{b-a}\int_a^b f(x)dx\) - average value of the rate function

Think of it this way: if \(f(t)\) represents speed at time \(t\), then:

  • \(\int_a^b f(t)dt\) is the total distance traveled
  • \(\frac{1}{b-a}\int_a^b f(t)dt\) is the average speed over the time period

The relationship is: Total Change = Average Rate × Time Interval

This is why \(\int_a^b f(x)dx = (b-a) \cdot \text{Average value of } f\)