Integration by Substitution: Business Calculus Study Notes

Chapter 5: Integration

Section 2: Integration by Substitution

Integration by substitution is a fundamental technique in calculus used to simplify integrals, especially those that are the result of the chain rule in differentiation. This method is essential for solving many business-related calculus problems, such as finding total cost, revenue, or price-demand relationships.

Review: The Chain Rule

• Chain Rule Definition: If a function is composed as f(g(x)), its derivative is $f'(g(x)) \cdot g'(x)$.

• Recognizing Chain Rule Forms: When an integrand appears as $E'[I(x)] \cdot I'(x)$, it is often the result of the chain rule and can be integrated by reversing the process.

• Application: Identifying chain rule forms allows for straightforward computation of antiderivatives.

Formulas for General Indefinite Integrals

• Indefinite Integral: The general form is $\int f'(g(x)) \cdot g'(x) \, dx = f(g(x)) + C$.

• Key Formulas: Recognizing integrands that fit standard formulas is crucial for efficient integration.

Integration by Substitution

When the integrand does not obviously fit a standard formula, substitution can be used to rewrite the integral in a simpler form.

• Purpose: To transform the integral into a form that is easier to evaluate.

• Method: Substitute $u = g(x)$, then $du = g'(x) dx$.

Definition: Differentials

• Differential dx: An arbitrary real number representing a small change in the independent variable.

• Differential dy: Defined as $dy = f'(x) dx$ for $y = f(x)$.

Examples of Differentials

• (A) $y = x^2$: $dy = 2x dx$

• (B) $u = e^{3x}$: $du = 3e^{3x} dx$

• (C) $w = \ln(4 + 5t)$: $dw = \frac{5}{4 + 5t} dt$ Additional info: The derivative of $\ln(4 + 5t)$ is $\frac{5}{4 + 5t}$.

Procedure: Integration by Substitution

1. Step 1: Select a substitution $u = g(x)$ that simplifies the integrand, aiming for $du$ to be a factor in the integrand.

2. Step 2: Rewrite the integrand entirely in terms of $u$ and $du$, eliminating $x$ and $dx$.

3. Step 3: Evaluate the new integral in terms of $u$.

4. Step 4: Express the antiderivative in terms of the original variable $x$.

Examples of Integration by Substitution

• Example 3: If $u = x^2 + 2x + 5$, then $du = (2x + 2) dx$.

• Example 4: Let $u = x^3 - 5$, then $du = 3x^2 dx$. Substitute and integrate, then revert to $x$.

• Example 5: Let $u = x^2 + 5$, then $du = 2x dx$. Substitute and integrate, then revert to $x$.

• Example 6: Let $u = x^3 - 5$, $du = 3x^2 dx$. If the integrand lacks the factor $3$, adjust by multiplying and dividing as needed.

• Example 7: Let $u = 4x^3$, $du = 12x^2 dx$. Adjust for missing factors in the integrand.

• Example 8: Let $u = 5 - 2x^2$, $du = -4x dx$. Adjust for missing negative factors.

• Example 9: Let $u = x + 6$, $du = dx$. Substitute and, if necessary, express $x$ in terms of $u$.

General Indefinite Integral Formulas

• Valid for $u$ as an independent variable or as a function of another variable, with $du$ as its differential.

• Common formulas include:

  • $\int f'(u) du = f(u) + C$

  • $\int u^n du = \frac{u^{n+1}}{n+1} + C$ (for $n \neq -1$)

  • $\int e^{u} du = e^{u} + C$

  • $\int \frac{1}{u} du = \ln|u| + C$

Business Application: Price-Demand Example

Integration by substitution is used in business calculus to solve real-world problems, such as finding price-supply equations from marginal price functions.

• Example 10: A company finds the marginal price $p'(x)$ at a supply level $x$ bottles per week. To find the price-supply equation $p(x)$, integrate $p'(x)$ using substitution.

• Substitution: Let $u = 3x + 25$, then $du = 3dx$.

• Application: Use the given supply and price data to solve for constants after integration.

Summary Table: Integration by Substitution Steps

Step	Description
1	Select substitution $u = g(x)$
2	Rewrite integrand in terms of $u$ and $du$
3	Integrate with respect to $u$
4	Express answer in terms of original variable $x$

Additional info: Adjusting for missing factors in $du$ is a common step in substitution. Multiply and divide by constants as needed to match the integrand to the differential.