Problem 49 - Calculus of a Single Variable

Question

Evaluating a Definite Integral In Exercises \(49-56\) , evaluate the definite integral. Use a graphing utility to verify your result. $$ \int_{0}^{4} \frac{5}{3 x+1} d x $$

Step-by-Step Solution

Answer

The evaluated definite integral is \(=\frac{5}{3} \ln 13\)

Step 1: Identify the formula for the integral

In this case, the function to integrate, \(\frac{5}{3x+1}\), is similar to the formula for the integral of \(\frac{1}{ax+b}\), which is \(\frac{1}{a} \ln |ax+b|\), \(a, b \in \mathbb{R}\) and \(a \neq 0\). Here, \(a=3\) and \(b=1\). So, \(\int \frac{1}{3x+1} d x = \frac{1}{3} \ln |3x+1|\) and multiple it by 5 which gives \(\int \frac{5}{3x+1} d x = \frac{5}{3} \ln |3x+1|\)

Step 2: Apply the fundamental theorem of calculus

The fundamental theorem of calculus states that \( \int_{a}^{b} f(x) d x = F(b) - F(a)\), where \(F(x)\) is the antiderivative of \(f(x)\). So, we apply it to our integral: \( \int_{0}^{4} \frac{5}{3x+1} d x = [\frac{5}{3} \ln |3(4)+1|] - [\frac{5}{3} \ln |3(0)+1|]\)

Step 3: Evaluate the definite integral

Substitute the limits \(0\) and \(4\) into the function: \(= \frac{5}{3} \ln |13| - \frac{5}{3} \ln |1|\). As the \(\ln(1)=0\) it simplifies to: \(=\frac{5}{3} \ln 13\)

Step 4: Verify the result using a graphing utility

Graph the function \(\frac{5}{3x+1}\) from \(0\) to \(4\) on a graphing utility and use the integral tool to find the definite integral. It should confirm the result \(=\frac{5}{3} \ln 13\)

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a critical link between differentiation and integration—the two main operations in calculus. It is often split into two parts. The first part tells us that if a function is continuous over an interval, then it has an antiderivative over that interval. This means you can "reverse" the process of differentiation by finding the antiderivative.

The second part states that if you have a continuous function, you can evaluate the definite integral of the function over an interval by using its antiderivative. In simpler terms:

• To find the definite integral of a function from point \(a\) to \(b\), find its antiderivative.
• Then compute the difference of the antiderivative evaluated at these endpoints \(F(b) - F(a)\).

In the context of the exercise, this theorem explains how we solved the integral \( \int_{0}^{4} \frac{5}{3x+1} \, dx \), by evaluating the antiderivative at the upper and lower bounds.

Antiderivative

An antiderivative is essentially the "inverse" of a derivative. Think of it like an undo button for differentiation. If you know the derivate of a function, the antiderivative tells you what the original function was before it was differentiated. For example, if the derivative of \( x^2 \) is \( 2x \), then \( x^2 \) would be an antiderivative of \( 2x \).

In solving definite integrals, we often first find an antiderivative. In our exercise, the function \( \frac{5}{3x+1} \) needed to have its antiderivative calculated. By using a known integral rule, we found

• The antiderivative of \( \frac{1}{3x+1} \) is \( \frac{1}{3} \ln |3x+1| \).
• Thus, multiplying by 5, the antiderivative of \( \frac{5}{3x+1} \) is \( \frac{5}{3} \ln |3x+1| \).

Finding antiderivatives correctly is crucial to evaluating definite integrals.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a fundamental mathematical function that often arises when dealing with exponential growth or decay problems, as well as in integrations involving rational functions such as \( \frac{1}{x} \). The natural logarithm has some properties that make it particularly useful in calculus:

• \( \ln(1) = 0 \), which simplifies many calculations.
• \( \ln(e) = 1 \), where \(e\) is the base of the natural logarithm, approximately equal to 2.71828.

In the given exercise, when computing the integral, the natural logarithm appears when we transform \( \int \frac{1}{3x+1} \, dx \) into \( \frac{1}{3} \ln |3x+1| \). The logarithmic nature means any constant factor in front of the function, like the "5" here, can be factored out in its antiderivative, leading to easier computations.

Graphing Utility

A graphing utility is a software tool or calculator that can graph functions and calculate integrals among other features. These utilities can be incredibly helpful in checking your work, as they provide a graphical view of the functions and verify computations of integrals.To solve the original exercise, after calculating the integral, a graphing utility can visually verify the accuracy of our solution:\(\int_{0}^{4} \frac{5}{3x+1} \, dx = \frac{5}{3} \ln 13\). Here's how:

• Load the function into the graphing utility.
• Set the window to show the function between the bounds 0 and 4.
• Use the integral feature to calculate the area under the curve—this should confirm previous calculations manually carried out.

By visualizing it, you ensure the solution is correctly computed and understand the integral's interpretation as an area under the curve.