Problem 2 - Calculus Single Variable

Question

In each of Exercises \(1-12,\) calculate the average value of the given function on the given interval. $$ f(x)=x^{2} \quad I=[3,7] $$

Step-by-Step Solution

Answer

The average value of the function \( f(x) = x^2 \) on the interval \([3,7]\) is \( \frac{79}{3} \).

Step 1: Understand the Formula for Average Value

The average value of a function \( f(x) \) on an interval \( [a, b] \) is given by the formula: \[ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \] In this exercise, our function \( f(x) = x^2 \) and the interval \( I = [3, 7] \). We need to apply this formula to find the average value.

Step 2: Set Up the Integral

We need to calculate the integral of \( f(x) = x^2 \) over the interval \( [3, 7] \). This is written as: \[ \int_{3}^{7} x^2 \, dx \] To find the average value, we will divide this result by \( 7 - 3 = 4 \).

Step 3: Calculate the Integral

Integrate \( x^2 \) over \( [3, 7] \). The antiderivative of \( x^2 \) is \( \frac{x^3}{3} \). Evaluate this from 3 to 7:\[ \left[ \frac{x^3}{3} \right]_{3}^{7} = \frac{7^3}{3} - \frac{3^3}{3} \] Calculate \( \frac{7^3}{3} = \frac{343}{3} \) and \( \frac{3^3}{3} = \frac{27}{3} = 9 \). Subtract to find the integral: \( \frac{343}{3} - 9 \).

Step 4: Simplify the Integral Result

First, convert 9 to an expression with denominator 3: \( 9 = \frac{27}{3} \). Now subtract these: \[ \frac{343}{3} - \frac{27}{3} = \frac{343 - 27}{3} = \frac{316}{3} \] This is the value of the integral over the interval.

Step 5: Calculate the Average Value

To find the average value, use the result of the integral, \( \frac{316}{3} \), and divide it by the length of the interval, 4: \[ \text{Average value} = \frac{1}{4} \times \frac{316}{3} = \frac{316}{12} \] Simplify \( \frac{316}{12} \): divide numerator and denominator by 4: \[ \frac{316}{12} = \frac{79}{3} \] This fraction cannot be simplified further.

Key Concepts

Definite Integral

A definite integral is a core concept in calculus that helps calculate the area under a curve, between two limits on a graph of a function. In our task, we are focused on finding the definite integral of the function \( f(x) = x^2 \) from 3 to 7. This process can be visualized as finding the area under the curve \( y = x^2 \) from \( x = 3 \) to \( x = 7 \).

• The definite integral is denoted as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration.
• For our example, \( a = 3 \) and \( b = 7 \), which are the start and end points of the interval on which we want to calculate the area.
• This integral calculates the total value accumulated by the function \( f(x) = x^2 \) within this interval.

Understanding the definite integral is crucial because it forms the basis for finding the average value of a function over an interval.

Antiderivative

An antiderivative, also known as an indefinite integral, is a function whose derivative is the original function we started with. In simpler terms, it's the reverse process of differentiation. For our function, \( f(x) = x^2 \), we need to find the antiderivative to compute its definite integral.

• The antiderivative of \( x^2 \) is \( \frac{x^3}{3} \). This means differentiating \( \frac{x^3}{3} \) will return \( x^2 \).
• This antiderivative is used to evaluate the definite integral. Once we find it, we plug the limits of the interval \( [3, 7] \) into the antiderivative to find the exact area \( \int_{3}^{7} x^2 \, dx \).
• To calculate, we find the difference: \[ \left[ \frac{x^3}{3} \right]_{3}^{7} = \frac{7^3}{3} - \frac{3^3}{3} \]

Understanding antiderivatives helps us evaluate definite integrals, and eventually find the average value of our function over a specified interval.

Interval Notation

Interval notation is a simple but essential concept in mathematics. It is a way to represent a range of numbers, especially useful in calculus to specify the domain over which we are interested in analyzing a function.

• In our case, the function \( f(x) = x^2 \) is examined over \( [3, 7] \).
• Brackets \( [3, 7] \) mean that both 3 and 7 are included in the interval, which are known as inclusive boundaries.
• This notation format helps us focus the definite integral and average value calculations on this specific interval, ensuring that only the portion of the function within these limits is considered.

Knowing how to properly read and write interval notation ensures that calculations are accurate and reflect the intended segment of the function's curve.

Mathematical Calculations

Mathematical calculations within this context involve substituting values, simplifying expressions, and understanding basic arithmetic.

• For the interval \( [3, 7] \), after computing the definite integral \( \frac{316}{3} \), we must find its average value over the given interval length by dividing by 4.
• This step involves simple division: \[ \text{Average value} = \frac{316}{3} \times \frac{1}{4} = \frac{316}{12} \]
• Simplifying this fraction is also part of the mathematical calculation, turning \( \frac{316}{12} \) into \( \frac{79}{3} \) after dividing both the numerator and the denominator by 4.

Accurate mathematical computations ensure that the final results, such as the average value, are correct and precisely represent the function's behavior over a specific interval.