Problem 15 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the indefinite integral. \(\int \frac{d x}{5-3 x}\)

Step-by-Step Solution

Answer

The indefinite integral is \(-\frac{1}{3} \ln |5 - 3x| + C\).

Step 1: Identify the Integral Form

The given integral is \( \int \frac{d x}{5-3 x} \). This is a form suitable for a simple substitution due to the linear expression \(5 - 3x\) in the denominator.

Step 2: Use U-Substitution

Let \( u = 5 - 3x \). Then, the derivative of \( u \) with respect to \( x \) is \( \frac{du}{dx} = -3 \). Consequently, \( dx = \frac{du}{-3} \). Substitute \( u \) and \( dx \) into the integral.

Step 3: Simplify the Integral

Substituting into the integral gives \( \int \frac{1}{u} \cdot \frac{du}{-3} = -\frac{1}{3} \int \frac{1}{u} \, du \). The factor of \(-\frac{1}{3}\) can be taken out of the integral.

Step 4: Integrate

The integral \( \int \frac{1}{u} \, du \) is a standard integral that evaluates to \( \ln |u| + C \), where \( C \) is the constant of integration. Therefore, \(-\frac{1}{3} \int \frac{1}{u} \, du = -\frac{1}{3} \ln |u| + C \).

Step 5: Substitute Back for x

Recall that \( u = 5 - 3x \). Substitute back to express the integral in terms of \( x \): \( -\frac{1}{3} \ln |5 - 3x| + C \).

Key Concepts

Understanding U-Substitution

U-substitution is a technique used to simplify the process of finding integrals, especially when dealing with complicated functions. It involves substituting a part of the integrand with a new variable, usually called "u," which stands for "substitution variable." In our original exercise, we have the integral \( \int \frac{d x}{5-3 x} \). Here, the expression \(5 - 3x\) can be defined as \(u\).

• Why use "u"? It's a new variable that simplifies the integration.
• How to choose "u"? Select a part of the expression that, when differentiated, relates to another part of the integrand, making substitution possible.

Once \(u = 5 - 3x\) is determined, differentiate \(u\) with respect to \(x\), giving \(\frac{du}{dx} = -3\), or rearranged as \(dx = \frac{du}{-3}\). This substitution simplifies our integral, making it easier to solve.

What is a Linear Expression?

A linear expression is an algebraic statement that involves variables raised to the power of one and added or subtracted together. They appear in the form \(ax + b\), where \(a\) and \(b\) are constants. In our problem, \(5 - 3x\) is a linear expression.

• Why are they simple? Because they only involve terms with degree one. No exponents or products of different variables.
• Application in integrals: Linear expressions in the denominator can often lead to natural logarithmic results in integration.

Using a linear expression helps in making the u-substitution straightforward and guides us toward finding the integral of more complex functions.

The Constant of Integration

The constant of integration, denoted as \(C\), is an arbitrary constant added to the integral's result.

• Why is it important? In indefinite integrals, \(C\) represents an infinite number of potential solutions.
• Indefinite vs. definite: Unlike definite integrals, indefinite integrals don't provide limits, thus requiring this constant to represent all possible equations.

When you integrate \( \int \frac{1}{u} \, du \), the result becomes \( \ln |u| + C \), showing how \(C\) completes the equation without affecting the derivative of the function.

Exploring the Natural Logarithm

The natural logarithm, usually denoted as \( \ln \), is a logarithm to the base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828. It's used extensively in calculus.

• Why "natural"? It naturally occurs in mathematical models of growth and decay.
• Connection to integration: The derivative of \( \ln |u| \) is \( \frac{1}{u} \), making it a common result when integrating functions of the form \( \frac{1}{u} \).

In our exercise, after substituting \(u = 5 - 3x\), we can find \( \int \frac{1}{u} \, du \) simplifying to \( \ln |u| \). This transformation shows how integrals of specific patterns result in natural logarithmic forms, which are prevalent due to the inverse relationship of exponentiation.