Problem 23 - Calculus Single Variable

Question

Compute the average value of \(f\) over \([a, b]\). \(f(x)=\sqrt{x}-1 / \sqrt{x} \quad a=1, b=4\)

Step-by-Step Solution

Answer

The average value of \( f(x) \) over \([1, 4]\) is \( \frac{8}{9} \).

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: \[ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx. \] In this problem, \( f(x) = \sqrt{x} - \frac{1}{\sqrt{x}} \) and the interval is \([1, 4]\).

Step 2: Setup the integral

Insert the function \( f(x) \) into the integral: \[ \int_{1}^{4} \left( \sqrt{x} - \frac{1}{\sqrt{x}} \right) \, dx. \] This integral represents the total area under the curve from \( x = 1 \) to \( x = 4 \).

Step 3: Simplify the integral into separate terms

The integral can be split into two components: \[ \int_{1}^{4} \sqrt{x} \, dx - \int_{1}^{4} \frac{1}{\sqrt{x}} \, dx. \] This separation allows us to evaluate each part individually.

Step 4: Evaluate the first integral

Compute \( \int_{1}^{4} \sqrt{x} \, dx \): Use the rule for the power of \( x \), \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( n = \frac{1}{2} \). Thus, the integral becomes: \[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2}. \]Evaluate from 1 to 4:\[ \frac{2}{3} [4^{3/2} - 1^{3/2}] = \frac{2}{3} [8 - 1] = \frac{14}{3}. \]

Step 5: Evaluate the second integral

Compute \( \int_{1}^{4} \frac{1}{\sqrt{x}} \, dx \): This is equivalent to \( \int x^{-1/2} \, dx \). Using the rule for the power of \( x \), the integral becomes:\[ \int x^{-1/2} \, dx = 2 x^{1/2}. \]Evaluate from 1 to 4:\[ 2 [4^{1/2} - 1^{1/2}] = 2 [2 - 1] = 2. \]

Step 6: Find the definite integral value

Combine the results from Steps 4 and 5:\[ \int_{1}^{4} \sqrt{x} \, dx - \int_{1}^{4} \frac{1}{\sqrt{x}} \, dx = \frac{14}{3} - 2 = \frac{8}{3}. \]

Step 7: Calculate the average value

Use the formula from Step 1 to find the average value:\[ \text{Average value} = \frac{1}{4-1} \cdot \frac{8}{3} = \frac{1}{3} \cdot \frac{8}{3} = \frac{8}{9}. \] The average value of \( f(x) \) over the interval \([1, 4]\) is \( \frac{8}{9} \).

Key Concepts

Understanding the Definite Integral

The definite integral is a fundamental concept in calculus that helps us calculate the area under a curve from one point to another. Unlike indefinite integrals, which represent families of functions and include a constant of integration, definite integrals provide the cumulative area between a curve and the x-axis over a specific interval. For functions like our given function, \[f(x) = \sqrt{x} - \frac{1}{\sqrt{x}},\]the definite integral evaluates the total signed area from the start of the interval \(a\) to the end \(b\). The notation \( \int_{a}^{b} f(x) \, dx \) not only symbolizes the integral you're performing but also bounds it with the specific limits. Here, those limits are the values that contain the interval \([1, 4]\).
It's important to understand that while calculating, any area below the x-axis will be subtracted in this total area calculation. This characteristic of definite integrals is useful in finding real-world applications, such as finding displacement when velocity is known across time.

Integral Evaluation Simplified

Calculating an integral involves breaking down complex expressions into more manageable pieces. When faced with an expression like \( \int_{a}^{b} \left( \sqrt{x} - \frac{1}{\sqrt{x}} \right) \, dx \), simplifying each term individually can greatly ease the process.

• First, we look at \( \sqrt{x} \), which can be thought of as \( x^{1/2} \). Using the power rule for integration, we integrate \( x^{1/2} \) to get \( \frac{2}{3}x^{3/2} \).
• Next, we tackle \( \frac{1}{\sqrt{x}} \), or \( x^{-1/2} \), which integrates to \( 2x^{1/2} \) using the same power rule.

This step-by-step breakdown using these rules not only simplifies the task but also strengthens understanding of how to deal with varying powers in integrals.
This integral evaluation is a crucial skill for any calculus student, allowing them to address problems more effectively.
Mastering this process can pave the way for more advanced calculus concepts in their studies.

Interval Calculation and Its Importance

Understanding intervals is key when working with functions and their integrals. Specifically, the interval \([a, b]\) dictates the start and end points over which you evaluate an integral. For our problem, the interval is \([1, 4]\), meaning we start calculating the area from \(x = 1\) to \(x = 4\).

• The length of the interval is calculated as \(b - a\). In our scenario, this becomes \(4 - 1 = 3\).
• This interval length is crucial when finding the average value of a function. It's used to divide the result of the definite integral calculation, ensuring the unit is correctly adjusted to reflect an average over that range.

In practical terms, interval calculation helps in ensuring that we focus on the right section of the graph. It's an essential skill for switching from the abstract concept of integrals to real-world applications like physics, where you might measure the average rate or quantity over time or distance.