Problem 56 - Calculus: Early Transcendentals

Question

Evaluate the definite integral. \( \displaystyle \int^3_0 \frac{dx}{5x + 1} \)

Step-by-Step Solution

Answer

The evaluated integral is \( \frac{1}{5} \ln(16) \).

Step 1: Recognize the Integral Form

Notice that the integral is of the form \( \int \frac{1}{ax + b} \, dx \), which is a logarithmic integral. The general solution for this form is \( \frac{1}{a} \ln|ax + b| + C \) when integrated indefinitely.

Step 2: Identify Parameter Values

Identify the parameters: \( a = 5 \) and \( b = 1 \). This information will be used to apply the formula for definite integrals.

Step 3: Apply the Logarithmic Integration Formula

Use the formula \( \int \frac{1}{5x + 1} \, dx = \frac{1}{5} \ln|5x + 1| + C \). This is the indefinite integral form which we'll evaluate over the given limits.

Step 4: Evaluate the Definite Integral

Compute the definite integral from 0 to 3: \[ \int_0^3 \frac{dx}{5x + 1} = \left[ \frac{1}{5} \ln|5x + 1| \right]_0^3 \].This means substituting the upper limit and the lower limit into the indefinite integral result.

Step 5: Substitute the Limits

Substitute the upper limit 3 into the expression: \[ \frac{1}{5} \ln|5 \times 3 + 1| = \frac{1}{5} \ln|16| \].Substitute the lower limit 0 into the expression:\[ \frac{1}{5} \ln|5 \times 0 + 1| = \frac{1}{5} \ln|1| \].

Step 6: Simplify the Expression

Since \( \ln(1) = 0 \), the expression simplifies to:\[ \frac{1}{5} ( \ln(16) - \ln(1) ) = \frac{1}{5} \ln(16) \].

Step 7: Final Result

The evaluated integral simplifies to the concise form:\[ \frac{1}{5} \ln(16) \]. This is the value of the definite integral from 0 to 3.

Key Concepts

Logarithmic Integral

A logarithmic integral involves an integral of the form \( \int \frac{1}{ax + b} \, dx \). This integral is called a *logarithmic* integral because its solution is related to the natural logarithm function. In calculus, when we solve an integral like this, we make use of a known formula:

• For \( \int \frac{1}{ax + b} \, dx \), the solution is \( \frac{1}{a} \ln|ax + b| + C \), where \( C \) represents the constant of integration.

This type of integral shows up frequently in problems involving rates of growth or decay. It's essential to understand how to apply the formula and adapt it to definite integrals, where the integration limits are specified. For definite integrals, we don't need to include the constant \( C \) because the evaluation over specified limits accounts for changes without it.
In our example, the function \( \frac{1}{5x + 1} \) does not inherently solve to a simple fraction, hence the logarithmic solution comes in to solve it by transforming it into a natural log form.

Integration Techniques

When encountering a calculus problem that involves integration, it is crucial to identify which integration technique to use. For a problem like \( \int^3_0 \frac{dx}{5x + 1} \), recognizing that it fits the logarithmic integral form is pivotal. Here are some fundamental steps for integrating such functions:

• **Identify the integral form**: Check if it matches a familiar structure like \( \int \frac{1}{ax + b} \).
• **Apply the appropriate formula**: Use the rule \( \frac{1}{a} \ln|ax + b| \) for indefinite integrals.
• **Evaluate limits for definite integrals**: In range [a,b], calculate using \( \left[ \frac{1}{a} \ln|ax + b| \right]_a^b \).

In the step-by-step solution, we first transform the integrable function using the logarithmic rule and then apply limits from 0 to 3. Integration techniques like this are not only a great way to simplify complex problems but also showcase how powerful logarithmic transformations can be in calculus.

Calculus Concepts

Calculus is a powerful mathematical tool used for analyzing changes. There are key concepts in calculus which help in understanding both continuity and the structure of integrals.
**Definite Integrals** are used to calculate the accumulation of a quantity over an interval. It differs from an indefinite integral, which expresses a family of functions. The definite integral is depicted by the symbol \( \,\int_a^b f(x) \, dx \), where \( a \) and \( b \) are the integration limits.
**Evaluation of Definite Integrals** involves calculating the exact value by substituting upper and lower limits into the antiderivative (or indefinite integral). For example, after computing \( \left[ \frac{1}{5} \ln|5x + 1| \right]_0^3 \), you plug in 3 to get the upper bound value and 0 for the lower bound, resulting in \( \frac{1}{5} \ln|16| - \frac{1}{5} \ln|1| \).
Understanding these calculus concepts enables students to solve integrals effectively and appreciate how they apply to real-world scenarios, such as calculating areas or solving physics problems involving motion.