MAC2233 (6) - Calculus for Management Unit 4 study guides

Key Concepts and Definitions

• Integration calculates the area under a curve, the opposite of differentiation which finds the slope of a tangent line at a point
• Definite integrals have specific start and end points (bounds) while indefinite integrals lack bounds and include a constant of integration (C)
  • Definite integrals are written as $\int_a^b f(x) dx$ where $a$ and $b$ are the lower and upper bounds
  • Indefinite integrals are written as $\int f(x) dx = F(x) + C$
• Antiderivatives are the family of functions whose derivative is the original function
  • Example: $\int x^2 dx = \frac{1}{3}x^3 + C$ because $\frac{d}{dx}(\frac{1}{3}x^3 + C) = x^2$
• The Fundamental Theorem of Calculus connects differentiation and integration, stating that definite integrals can be evaluated using antiderivatives
• Integration techniques include substitution, integration by parts, and partial fractions which help solve more complex integrals
• Applications of integration in business and economics include calculating consumer and producer surplus, present value of continuous income streams, and future value of investments

Basic Integration Rules

• The power rule for integration states that $\int x^n dx = \frac{1}{n+1}x^{n+1} + C$ for $n \neq -1$
  • Example: $\int x^3 dx = \frac{1}{4}x^4 + C$
• The constant multiple rule allows constants to be factored out of the integral: $\int kf(x) dx = k\int f(x) dx$
• The sum and difference rules state that the integral of a sum or difference is the sum or difference of the integrals: $\int [f(x) \pm g(x)] dx = \int f(x) dx \pm \int g(x) dx$
• Integrate common functions using the following rules:
  • $\int e^x dx = e^x + C$
  • $\int \ln(x) dx = x\ln(x) - x + C$
  • $\int \sin(x) dx = -\cos(x) + C$
  • $\int \cos(x) dx = \sin(x) + C$
• The net signed area can be found using definite integrals, with areas below the x-axis counted as negative

Substitution Method

• The substitution method is used when an integral contains a function and its derivative, allowing for a simpler expression to be integrated
• To use substitution, identify a part of the integral that can be replaced with a new variable (often denoted as $u$)
  • The substituted part should be a composite function, and the remaining part should include its derivative
• Rewrite the integral in terms of $u$, replace $dx$ with $du$ based on the relationship between $u$ and $x$
• Integrate the new expression with respect to $u$, then substitute back the original variable to obtain the antiderivative in terms of $x$
• Common substitutions include:
  • Expressions involving $\sqrt{a^2 - x^2}$ often use $u = \sin(\theta)$ or $u = \cos(\theta)$
  • Expressions with $\sqrt{x^2 + a^2}$ often use $u = \tan(\theta)$
  • Expressions with $\sqrt{x^2 - a^2}$ often use $u = \sec(\theta)$

Integration by Parts

• Integration by parts is a technique used when an integral contains a product of functions, one of which is easier to integrate than the other
• The formula for integration by parts is $\int u dv = uv - \int v du$
  • Choose $u$ as the function that is easier to differentiate and $dv$ as the function that is easier to integrate
• Compute $du$ by differentiating $u$ and $v$ by integrating $dv$
• Substitute the expressions for $u$, $du$, $v$, and $dv$ into the integration by parts formula and simplify
• Repeat the process if the resulting integral is still difficult to solve, using integration by parts multiple times if necessary
• LIATE is a mnemonic to help choose $u$ in order of preference: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential functions

Partial Fractions

• Partial fraction decomposition is a technique used to integrate rational functions (quotients of polynomials) by splitting them into simpler fractions
• Proper rational functions (degree of numerator < degree of denominator) are required for partial fraction decomposition
• Factor the denominator completely, then set up the partial fraction decomposition based on the factors:
  • Linear factors $(ax + b)$ correspond to terms of the form $\frac{A}{ax + b}$
  • Repeated linear factors $(ax + b)^n$ correspond to terms of the form $\frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \cdots + \frac{A_n}{(ax + b)^n}$
  • Quadratic factors $(ax^2 + bx + c)$ correspond to terms of the form $\frac{Bx + C}{ax^2 + bx + c}$
• Solve for the unknown coefficients $(A, B, C)$ by equating the original rational function to the partial fraction decomposition and comparing coefficients or evaluating at specific points
• Integrate each resulting fraction using basic integration rules or substitution

Applications in Business and Economics

• Consumer and producer surplus can be calculated using definite integrals
  • Consumer surplus is the area below the demand curve and above the market price
  • Producer surplus is the area above the supply curve and below the market price
• Present value of continuous income streams can be found using integrals
  • The present value of a continuous income stream $f(t)$ from time $a$ to $b$ with a discount rate $r$ is given by $PV = \int_a^b f(t)e^{-rt} dt$
• Future value of investments with continuous compounding can be calculated using the formula $FV = PV \cdot e^{rt}$
  • $PV$ is the present value, $r$ is the annual interest rate, and $t$ is the number of years
• Lorenz curves and Gini coefficients, which measure income inequality, can be computed using integrals
  • The Lorenz curve plots the cumulative percentage of income earned by the bottom x% of the population
  • The Gini coefficient is the ratio of the area between the Lorenz curve and the line of perfect equality to the total area under the line of perfect equality

Common Mistakes and How to Avoid Them

• Forgetting to add the constant of integration (C) when finding indefinite integrals
  • Always include +C when finding antiderivatives, unless given initial conditions
• Incorrectly applying the power rule by forgetting to add 1 to the exponent and divide by the new exponent
  • Remember that $\int x^n dx = \frac{1}{n+1}x^{n+1} + C$, not $\frac{1}{n}x^n + C$
• Misusing the substitution method by choosing an inappropriate substitution or forgetting to change the differential (dx) to match the substitution
  • Ensure that the substituted part includes both the function and its derivative
  • Change the differential to match the substitution, e.g., if $u = 2x$, then $du = 2dx$ or $dx = \frac{1}{2}du$
• Incorrectly applying the integration by parts formula, especially when using it multiple times
  • Keep track of the signs and the order of the terms in the formula: $\int u dv = uv - \int v du$
  • Be careful when choosing u and dv, as an incorrect choice can lead to a more complicated integral
• Making algebraic errors when solving for coefficients in partial fraction decomposition
  • Double-check your algebra and simplify your work to avoid errors
  • Verify that your decomposition is correct by adding the fractions and comparing to the original rational function

Practice Problems and Solutions

1. Evaluate $\int (3x^2 + 2x - 1) dx$ Solution: $\int (3x^2 + 2x - 1) dx = x^3 + x^2 - x + C$

2. Find $\int \frac{x+2}{\sqrt{x}} dx$ using substitution Solution: Let $u = \sqrt{x}$, then $du = \frac{1}{2\sqrt{x}}dx$ or $dx = 2u du$ $\int \frac{x+2}{\sqrt{x}} dx = \int \frac{u^2+2}{u} \cdot 2u du = 2\int (u^2 + 2) du = 2(\frac{1}{3}u^3 + 2u) + C$ Substituting back $u = \sqrt{x}$, we get $\frac{2}{3}x^{3/2} + 4\sqrt{x} + C$

3. Evaluate $\int x\ln(x) dx$ using integration by parts Solution: Let $u = \ln(x)$ and $dv = x dx$ Then $du = \frac{1}{x}dx$ and $v = \frac{1}{2}x^2$ $\int x\ln(x) dx = \frac{1}{2}x^2\ln(x) - \int \frac{1}{2}x^2 \cdot \frac{1}{x}dx = \frac{1}{2}x^2\ln(x) - \frac{1}{4}x^2 + C$

4. Integrate $\int \frac{2x+3}{(x+1)(x-2)} dx$ using partial fractions Solution: $\frac{2x+3}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2}$ Solving for A and B: $2x+3 = A(x-2) + B(x+1)$ Substitute x = -1: $1 = -3A$, so $A = -\frac{1}{3}$ Substitute x = 2: $7 = 3B$, so $B = \frac{7}{3}$ $\int \frac{2x+3}{(x+1)(x-2)} dx = \int (\frac{-1/3}{x+1} + \frac{7/3}{x-2}) dx = -\frac{1}{3}\ln|x+1| + \frac{7}{3}\ln|x-2| + C$

5. Calculate the present value of a continuous income stream $f(t) = 1000e^{0.02t}$ from t = 0 to t = 10 years, with a discount rate of 5% per year Solution: $PV = \int_0^{10} 1000e^{0.02t}e^{-0.05t} dt = 1000\int_0^{10} e^{-0.03t} dt$ = 1000 \cdot \frac{-1}{0.03}e^{-0.03t}|_0^{10} = \frac{1000}{0.03}(1 - e^{-0.3}) \approx \24,617.97$