Textbook Readings: 1.4 Integration Formulas and the Net Change Theorem (Calculus 2)

Integration Formulas and the Net Change Theorem

Overview and Learning Objectives

Basic integration formulas are essential for solving key applied problems across physics, engineering, and mathematics. Integration formulas are typically presented in terms of indefinite integrals, but their application frequently extends to definite integrals depending on the context of the problem.

Key Learning Objectives:

1. Apply basic integration formulas to evaluate definite and indefinite integrals.

2. Explain the conceptual and physical significance of the Net Change Theorem.

3. Use the Net Change Theorem to compute net displacement, total distance, and fluid consumption in real-world scenarios.

4. Utilize the symmetry properties of even and odd functions to evaluate integrals over symmetric intervals.

Definite vs. Indefinite Integrals

• Definite Integral: Represents a specific numerical value (when the limits of integration are constants) or a single function (when one or both limits of integration are variables). It computes signed areas under curves.

• Indefinite Integral: Represents a whole family of antiderivative functions, all differing by an arbitrary constant $C$.

Basic Integration Formulas

Basic antiderivative rules allow functions to be rewritten and simplified prior to integration.

Example 1.23: Integrating a Function Using the Power Rule

Problem: Use the power rule to integrate the function: $\int_1^4 \sqrt{t}(1 + t)\,dt$

Solution:

First, rewrite the radical expression using fractional exponents and distribute across the binomial to put the integrand into a form suitable for the power rule: $\int_1^4 \sqrt{t}(1 + t)\,dt = \int_1^4 t^{1/2}(1 + t)\,dt = \int_1^4 (t^{1/2} + t^{3/2})\,dt$

Next, apply the power rule for integration, $\int t^n\,dt = \frac{t^{n+1}}{n+1}$: $\int_1^4 (t^{1/2} + t^{3/2})\,dt = \left[ \frac{2}{3}t^{3/2} + \frac{2}{5}t^{5/2} \right]_1^4$

Now, evaluate the antiderivative at the upper limit $t = 4$ and lower limit $t = 1$: $\left[ \frac{2}{3}(4)^{3/2} + \frac{2}{5}(4)^{5/2} \right] - \left[ \frac{2}{3}(1)^{3/2} + \frac{2}{5}(1)^{5/2} \right]$ $= \left[ \frac{2}{3}(8) + \frac{2}{5}(32) \right] - \left[ \frac{2}{3}(1) + \frac{2}{5}(1) \right]$ $= \left[ \frac{16}{3} + \frac{64}{5} \right] - \left[ \frac{2}{3} + \frac{2}{5} \right]$ $= \left( \frac{80 + 192}{15} \right) - \left( \frac{10 + 6}{15} \right)$ $= \frac{272}{15} - \frac{16}{15} = \frac{256}{15}$

Checkpoint 1.21

Problem: Find the definite integral of $f(x) = x^2 - 3x$ over the interval $[1, 3]$.

Solution: $\int_1^3 (x^2 - 3x)\,dx = \left[ \frac{x^3}{3} - \frac{3x^2}{2} \right]_1^3$ $= \left( \frac{3^3}{3} - \frac{3(3)^2}{2} \right) - \left( \frac{1^3}{3} - \frac{3(1)^2}{2} \right)$ $= \left( 9 - \frac{27}{2} \right) - \left( \frac{1}{3} - \frac{3}{2} \right)$ $= \left( -\frac{9}{2} \right) - \left( -\frac{7}{6} \right)$ $= -\frac{27}{6} + \frac{7}{6} = -\frac{20}{6} = -\frac{10}{3}$

The Net Change Theorem

The Net Change Theorem considers the integral of a rate of change. It states that when a quantity changes, its final value equals its initial value plus the integral of its rate of change.

Formal Statement (Theorem 1.6)

The new value of a changing quantity equals the initial value plus the integral of the rate of change: $F(b) = F(a) + \int_a^b F'(x)\,dx$

Equivalently, subtracting $F(a)$ from both sides gives the net change formula: $\int_a^b F'(x)\,dx = F(b) - F(a)$

Applications and Significance

• Net change applies to displacement, fluid dynamics, total production, work, and heat transfer.

• Net change automatically accounts for negative rate values (such as motion in reverse or fluid drainage) without requiring multiple integrals.

Motion Applications: Net Displacement vs. Total Distance

When applying the Net Change Theorem to a velocity function $v(t)$, the result represents net displacement, which takes direction into account.

Conceptual Example: Car Traveled North and South

Suppose a car moves north (positive direction) at $40\,\text{mph}$ from 2 p.m. to 4 p.m. ($t = 2$ to $t = 4$), and then moves south at $30\,\text{mph}$ from 4 p.m. to 5 p.m. ($t = 4$ to $t = 5$).

• Net Displacement: Integrates the velocity function $v(t)$ directly:   $\int_2^5 v(t)\,dt = \int_2^4 40\,dt + \int_4^5 -30\,dt = 80 - 30 = 50$   At 5 p.m., the car is $50\,\text{mi}$ north of its initial 2 p.m. position.

• Total Distance Traveled: Integrates the absolute value of velocity $|v(t)|$:   $\int_2^5 |v(t)|\,dt = \int_2^4 40\,dt + \int_4^5 30\,dt = 80 + 30 = 110$   The total distance driven by the car is $110\,\text{mi}$.

Mathematical Definitions for Motion:

• Net Displacement: $\int_a^b v(t)\,dt$

• Total Distance Traveled: $\int_a^b |v(t)|\,dt$

Example 1.24: Finding Net Displacement

Problem: Given a velocity function $v(t) = 3t - 5\,\text{m/s}$ for a particle in motion from time $t = 0$ to time $t = 3$, find the net displacement of the particle.

Solution: Apply the Net Change Theorem: $\int_0^3 (3t - 5)\,dt = \left[ \frac{3t^2}{2} - 5t \right]_0^3$ $= \left[ \frac{3(3)^2}{2} - 5(3) \right] - 0$ $= \frac{27}{2} - 15 = \frac{27}{2} - \frac{30}{2} = -\frac{3}{2}$

The net displacement is $-\frac{3}{2}\,\text{m}$.

Example 1.25: Finding Total Distance Traveled

Problem: Use the velocity function from Example 1.24 ($v(t) = 3t - 5\,\text{m/s}$) to find the total distance traveled over the time interval $[0, 3]$.

Practical Applications of the Net Change Theorem

Motion and Fluid Applications

• Total Distance Traveled: Integrated using the absolute value of velocity, splitting the integral where $v(t) = 0$:   $\int_a^b |v(t)|\,dt$

  • Linear Example: For $v(t) = 3t - 5$ on $[0, 3]$, split at $t = \frac{5}{3}$ to find total distance = $\frac{41}{6}\,\text{m}$.

  • Exponential Example (Checkpoint 1.22): For $f(t) = \frac{1}{2}e^t - 2$ on $[0, 2]$:

  • Net Displacement: $\int_0^2 \left(\frac{1}{2}e^t - 2\right)\,dt = \frac{1}{2}e^2 - \frac{9}{2}\,\text{m}$

  • Total Distance: Split at $t = \ln(4)$ to get $4\ln(4) + \frac{1}{2}e^2 - \frac{15}{2}\,\text{m}$

• Fluid Consumption:

  • Example 1.26: Consumption rate $5 - 0.1t^3\,\text{gal/hr}$ over $[0, 2]$ yields $\int_0^2 (5 - 0.1t^3)\,dt = 9.6\,\text{gal}$.

• Piecewise Rate Functions: Integrate each subinterval with its corresponding rate function.

  • Iceboat Examples: Distance covered at speed $2v(t)$ over $[0, 1]$ equals $25\,\text{mi}$ when wind stays steady (Example 1.27), and $17.5\,\text{mi}$ when wind dies down (Checkpoint 1.23).

Integrating Even and Odd Functions

For continuous functions integrated over symmetric intervals $[-a, a]$:

• Even Functions: $f(-x) = f(x)$ (symmetric about $y$-axis)

  • Rule: $\int_{-a}^a f(x)\,dx = 2 \int_0^a f(x)\,dx$

  • Examples: $\int_{-2}^2 (3x^8 - 2)\,dx = \frac{1000}{3}$ and \int_{-2}^2 x^4\,\dx = \frac{64}{5}

• Odd Functions: $f(-x) = -f(x)$ (symmetric about origin)

  • Rule: $\int_{-a}^a f(x)\,dx = 0$

  • Example: $\int_{-\pi}^\pi -5\sin(x)\,dx = 0$