Problem 93 - Calculus and its applications

Question

Evaluate. $$ \int \frac{d x}{a x+b} $$

Step-by-Step Solution

Answer

\( \int \frac{d x}{a x+b} = \frac{1}{a} \ln |a x + b| + C \)

Step 1: Identify the Form of the Integral

The integral \( \int \frac{d x}{a x + b} \) is a basic integral similar to \( \int \frac{d x}{u} \) for some linear function \( u = a x + b \). This form suggests a substitution method might simplify the integral.

Step 2: Substitution

Perform the substitution \( u = a x + b \), which implies that \( du = a \, dx \). Therefore, \( dx = \frac{du}{a} \). Substituting for \( dx \) in the integral gives \[ \int \frac{1}{u} \cdot \frac{du}{a} = \frac{1}{a} \int \frac{du}{u}. \]

Step 3: Integrate with Respect to u

The integral \( \int \frac{1}{u} \, du \) is a standard integral that equals \( \ln |u| + C \) where \( C \) is the constant of integration. Thus, we have \[ \frac{1}{a} \ln |u| + C. \]

Step 4: Back-substitute for x

Substitute back \( u = a x + b \) to express the result in terms of \( x \). This gives us the final integrated result: \[ \frac{1}{a} \ln |a x + b| + C. \]

Key Concepts

Definite and Indefinite Integrals

In calculus, both definite and indefinite integrals play crucial roles. While an indefinite integral is the most general form of antiderivatives and comes with a constant of integration, a definite integral provides a specific value representing the area under the curve of a given function. Indefinite integrals are expressed as \( \int f(x) \, dx \) plus a constant \( C \), which accounts for any vertical shifts:

• Indefinite integrals do not have limits.
• The presence of \( C \) ensures all possible antiderivatives are included.

When handling definite integrals, the integration is carried out between prescribed upper and lower limits. These limits indicate the interval over which the area is calculated:

• Denoted as \( \int_{a}^{b} f(x) \, dx \).
• No constant \( C \) is necessary since it cancels out when computing the definite value.

It's essential to comprehend these differences, as they help in solving problems appropriately depending on the situation—whether you need a specific value or a general function.

Integration Techniques

There are various techniques that can simplify the process of integration, allowing for easier calculation of antiderivatives. Integrating directly from the definition can be challenging, so we use techniques such as substitution and integration by parts:

• Basic Techniques: involves recognizing standard forms and using basic antiderivatives.
• Substitution Method: simplifies integrals by transforming variables and making the integral easier to solve.
• Integration by Parts: applies when the integral is a product, using the formula \( \int u \, dv = uv - \int v \, du \).

These methods are valuable tools in a calculus student's arsenal, enabling the integration of complex functions which otherwise seem daunting. Familiarity with these techniques is crucial for efficiently tackling integrals in homework and exams.

Substitution Method

The substitution method, a favorite among students, simplifies integrals by making a strategic change of variables. In this method, a variable \( u \) replaces a portion of the original function, often making the integral more manageable:

• Identify a "inner" function: Look for a part of the integrand that can be set as \( u \).
• Calculate \( du \): Find the derivative of \( u \) with respect to the original variable, \( x \).
• Substitute \( dx \) with \( du \): This grows from \( dx = \frac{du}{something} \).

For the exercise \( \int \frac{d x}{a x + b} \), you can set \( u = a x + b \). This transforms \( dx = \frac{du}{a} \), leading to the integral \( \frac{1}{a} \int \frac{du}{u} \). Finally, solve the integral in terms of \( u \) and revert back to \( x \). The substitution method not only simplifies calculations but also enhances understanding of underlying relationships in functions.