University Mathematics Workshop: Advanced Integration and Optimization Notes

Integration Techniques: Definite Integrals and Substitution

• Simplification and Substitution Steps     - In the final stages of an integration problem, coefficients should be simplified. For example, a coefficient of $3/3$ is canceled, and a term involving $4$ may be reduced to $\frac{1}{2}$. Similarly, an expression like $\frac{15}{4}$ may become $\frac{15}{2}$ during simplification.     - When using integration by substitution (u-substitution), the variable $u$ must be replaced with the original expression in terms of $x$ before applying limits (unless the limits themselves were converted).     - An example of a back-substituted expression is:       $= \frac{1}{2} (2x - 5)^{\frac{3}{2}} + \frac{15}{2} (2x - 5)^{\frac{1}{2}}$     - Once the variables are back in terms of $x$, the limits of integration (in this case, $3$ and $7$) are placed outside big square brackets to denote the final evaluation step.

• Strategies for Limits of Integration     - There are two primary options for handling limits during substitution:         1. Option 1 (Recommended): Drop the limits of integration entirely during the intermediate working steps. Complete all integration and back-substitute the original $x$ values into the formula. Finally, apply the original limits of integration. This method reduces the odds of getting mixed up.         2. Option 2: Convert the limits of integration. Plug the original $x$ values (e.g., $x = 3$ and $x = 7$) into the substitution formula for $u$ to find the corresponding $u$ values. Use these new values as the limits of integration without back-substituting for $x$.     - While both methods are mathematically valid, Option 1 is recommended for students to avoid confusion, especially when comparing work to a memorandum where limits may have changed.

• Common Error Sources in Integration     - Even after correctly identifying the integral, students often fail to reach the final numerical answer (in one example, the answer was $28$) due to:         - Misplacing minus signs during subtraction of the upper and lower bounds.         - Errors in calculator input.     - Calculus Classification: Questions involving rearrangements (e.g., solving for $x$ because the derivative of the inside part doesn't automatically cancel) are categorized as more difficult (e.g., "EQ1"). This makes them "not nice" for standard integration by substitution.     - Integration by Parts Constraints: This method is not recommended for algebraic-only problems where the LIATE (Logarithmic, Inverse Trigonometric, Algebraic, Trigonometric, Exponential) rule does not provide a clear distinction (e.g., both parts are algebraic).

Constrained Optimization: Maximizing Profit

• Problem Context     - The goal is to determine the number of items (e.g., types of lamps) to produce to maximize profit ($P$).     - The profit formula often contains two variables ($x$ and $y$). To differentiate, one variable must be eliminated using a constraint equation.

• Methodology: Substitution vs. Implicit Differentiation     - Some students attempt implicit differentiation ($\frac{dP}{dx}$), but common mistakes include using $\frac{dy}{dx}$ incorrectly or failing to account for multiple instances of variables.     - The Recommended Approach: Rearrange the constraint equation and substitute it into the profit formula.         - Example Constraint: $x + 2y = 80$         - Rearrange for $x$: $x = 80 - 2y$         - Rearrange for $y$: $y = 40 - \frac{x}{2}$     - This allows the creation of a profit formula solely in terms of $x$.

• Algebraic Hazards     - Simplifying the profit formula requires extreme care with signs, especially when squaring terms and distributing negative signs.     - Example simplification sequence:         1. Start with: $P = 60x + 100(40 - \frac{x}{2}) - x^2 - y^2$ (Hypothetical placeholder based on transcript context).         2. Simplified terms: $60x + 4000 - 50x - x^2$         3. Squaring the substitution: $(40 - \frac{x}{2})^2 = 1600 - 40x + \frac{x^2}{4}$         4. Applying a negative to the square: $-(1600 - 40x + \frac{x^2}{4}) = -1600 + 40x - \frac{x^2}{4}$     - Combining terms: $50x + 2400 - x^2 - \frac{x^2}{4} = 50x + 2400 - \frac{5}{4}x^2$

• Finding the Optimal Production Level     - Once the formula is in one variable, find the derivative and set it to zero.     - $\frac{dP}{dx} = 50 - \frac{10}{4}x = 50 - \frac{5}{2}x$     - Setting the derivative to zero: $50 = \frac{5}{2}x \rightarrow 100 = 5x \rightarrow x = 20$     - Back-substitute into the constraint: $y = 40 - \frac{20}{2} = 30$

• Key Terms and Operations     - Constraint: Signal to rearrange and plug into the other formula.     - Elasticity: If mentioned, set elasticity to $1$ to maximize (do not differentiate equals zero).     - Related Rates: If the word "rates" is used, use one derivative to find another derivative.

Financial Mathematics: Interest Components and Compounding

• Interest Component of a Specific Payment     - To find the interest component of a specific payment (e.g., the $300^{th}$ payment), one must understand that while the regular monthly payment ($PMT$) is constant, the ratio of interest to principal changes.     - Example Calculation:         1. Identify the constant payment: $R20,140.71$         2. Calculate the Balance Outstanding ($BO$) with one month remaining. This is done by moving the final payment backwards in time by one month using the Present Value ($PV$) calculation.         3. $PV = \frac{FV}{(1 + i)^n}$         4. For the last payment: $PV = \frac{20,140.71}{1 + (\frac{0.09}{12})^1}$         5. This results in a balance outstanding of approximately $R19,990.78$.         6. The difference between the payment amount ($R20,140.71$) and the balance at the start of the month ($R19,990.78$) represents the interest accumulated that month.

• Continuous Compounding and Effective Rates     - Continuous Compounding Formulas:         - Present Value: $PV = FV e^{-rt}$         - Future Value: $FV = PV e^{rt}$     - Effective Interest Rate ($r_{eff}$):         - For discrete compounding: $r_{eff} = (1 + \frac{r}{m})^m - 1$, where $m$ is the number of compounding periods.         - For continuous compounding: $r_{eff} = e^r - 1$. Continuous compounding implies an infinite number of compounding periods ($m \rightarrow \infty$).

Advanced Integration: Inverse Trigonometric Substitution

• Problem Scenario     - Evaluate: $\int_0^{\frac{1}{2}} \frac{2\arccos(x)}{\sqrt{1 - x^2}} dx$     - Strategy Determination:         - Integration by parts is ruled out because the "inside part" of the square root is non-linear ($1 - x^2$).         - Rule of thumb: Look for a part of the integrand whose derivative matches another part.     - The Substitution:         - Let $u = \arccos(x)$         - The derivative $\frac{du}{dx} = -\frac{1}{\sqrt{1 - x^2}}$         - Therefore, $dx = -\sqrt{1 - x^2} du$     - Integration Process:         1. Move the constant $2$ outside the integral.         2. Multiply the terms: $2 \int u \times \frac{1}{\sqrt{1 - x^2}} \times (-\sqrt{1 - x^2}) du$         3. Cancellation yields: $-2 \int u du$         4. The result of the integral is $-u^2$.     - Evaluation:         - Replace $u$ with $\arccos(x)$, resulting in $-(\arccos(x))^2$.         - Apply limits $0$ and $\frac{1}{2}$.         - Boundary values: $\arccos(\frac{1}{2}) = \frac{\pi}{3}$ and $\arccos(0) = \frac{\pi}{2}$.         - Final calculation must be done in Radians on the calculator.

Questions & Discussion

• Question regarding integration by parts for the $\arccos$ problem:     - Response: While parts might be attempted, it is often ineffective because the right-hand side of the integration by parts formula might result in an integral that is not simpler than the original, leading to an infinite loop.
• Question regarding handling limits in integration by parts:     - Response: The cleanest way is to perform the integration without bounds first, arrive at the final expression (without the $+c$), and then apply the original bounds in square brackets.
• Technical Discussion:     - Students mentioned difficulty accessing materials on specific devices (iPhone/iPad). The speaker recommended using Google Chrome on a lab computer for better compatibility.
• Question on simple interest:     - Response: While simple interest is unlikely to appear as a major exam question, students should be prepared for everything. Usually, financial math focuses on continuous compounding and annuities.