Problem 62 - Applied Calculus

Question

Find the exact average value of \(f(x)=1 /(x+1)\) on the interval \(x=0\) to \(x=2 .\) Sketch a graph showing the function and the average value.

Step-by-Step Solution

Answer

The average value is \(\frac{\ln(3)}{2}\).

Step 1: Identify the Formula for Average Value

The average value of a function \(f(x)\) over the interval \([a, b]\) is given by the formula: \[ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]. For our exercise, \(a = 0\), \(b = 2\), and \(f(x) = \frac{1}{x+1}\).

Step 2: Set Up the Integral

Using the formula, set up the integral that needs to be calculated: \[ \text{Average value} = \frac{1}{2 - 0} \int_{0}^{2} \frac{1}{x+1} \, dx \]. Simplify this to: \[ \text{Average value} = \frac{1}{2} \int_{0}^{2} \frac{1}{x+1} \, dx \].

Step 3: Compute the Integral

The integral \(\int \frac{1}{x+1} \, dx\) is a standard form, and its antiderivative is \(\ln|x+1| + C\). Compute the definite integral: \[ \int_{0}^{2} \frac{1}{x+1} \, dx = \left[ \ln|x+1| \right]_{0}^{2} = \ln(3) - \ln(1) = \ln(3). \]

Step 4: Calculate the Average Value

Substitute the result of the integral into the formula for the average value: \[ \text{Average value} = \frac{1}{2} \times \ln(3) = \frac{\ln(3)}{2}. \]

Step 5: Sketch the Graph

Draw the graph of \(f(x) = \frac{1}{x+1}\) on the interval from \(x=0\) to \(x=2\). On the same graph, draw a horizontal line at the height of the average value, \(y = \frac{\ln(3)}{2}\), showing the average value of the function over this interval.

Key Concepts

Definite Integral

The definite integral is a fundamental concept in calculus. It allows us to calculate the accumulation of a quantity over an interval. When you integrate a function over a certain range, you are essentially summing up all infinitesimally small areas under the curve of that function. This helps in determining the total area between the graph of the function and the x-axis over the specified interval.

In our exercise, when we set up the integral to find the average value of the function, we used:

• The interval from 0 to 2, where 0 is the lower bound (a) and 2 is the upper bound (b).
• The function to integrate was \(f(x) = \frac{1}{x+1}\).
• The integral became \( \int_{0}^{2} \frac{1}{x+1} \, dx \).

This integral represents the total area under the curve of \(\frac{1}{x+1}\) from \(x = 0\) to \(x = 2\), giving us a precise measure of this accumulated area.

Antiderivative

The antiderivative of a function is another function whose derivative is the original function. It's the reverse process of differentiation. In the problem we're looking at, finding the antiderivative is essential for computing the definite integral.

For the function \(f(x) = \frac{1}{x+1}\), the antiderivative is known as:

• \( \ln|x+1| + C \)

where \(C\) is the constant of integration. This antiderivative tells us that if we differentiate \( \ln|x+1| \), we arrive back at \( \frac{1}{x+1} \). However, in the context of definite integrals, we focus only on the changes in indefinite terms, so the constant \(C\) doesn't affect the result.

In the solution, we evaluated this antiderivative from \(0\) to \(2\) to find the change, which helped us compute:

• \( \ln(3) - \ln(1) = \ln(3) \)

This result played a critical role in determining the average value of the function.

Graph of a Function

Drawing the graph of a function is a powerful way to visualize how a function behaves over an interval. In our exercise, the function \(f(x) = \frac{1}{x+1}\) was sketched between \(x = 0\) and \(x = 2\). This specific function is a decreasing function as \(x\) increases.

To find the average value visually, we also drew a horizontal line at \(y = \frac{\ln(3)}{2}\). This line represents the average height of the function over the given interval. The average value can sometimes be thought of as a height that, if kept constant over the interval, would result in the same total area under the curve as the varying heights of \(f(x)\) do.

Key points about sketching such graphs include:

• Identifying important points, such as the function value at boundaries.
• Understanding the behavior of the graph, noticing at \(x=0\), \(f(x)\) is \(1\), and it decreases as \(x\) approaches 2.
• The horizontal line helps visually confirm the calculated average value aligns with your expectations of the area's balance.

Sketching the graph is both a computational aid and a confirmation tool for understanding the mathematical concepts.