Problem 9 - Calculus Single Variable

Question

In each of Exercises \(1-12,\) calculate the average value of the given function on the given interval. $$ f(x)=60 / x^{2} \quad I=[1,3] $$

Step-by-Step Solution

Answer

The average value of the function is 20.

Step 1: Understanding the Exercise

We are asked to find the average value of the function \( f(x) = \frac{60}{x^2} \) over the interval \( [1, 3] \). The formula for the average value \( f_{avg} \) of a function \( f(x) \) on an interval \( [a, b] \) is \[ f_{avg} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx. \]

Step 2: Setting Up the Integral

Substitute \( f(x) = \frac{60}{x^2} \) and the interval \( [1, 3] \) into the average value formula: \[ f_{avg} = \frac{1}{3-1} \int_{1}^{3} \frac{60}{x^2} \, dx. \] Simplify the fraction: \[ f_{avg} = \frac{1}{2} \int_{1}^{3} \frac{60}{x^2} \, dx. \]

Step 3: Calculating the Integral

To solve the integral, \( \int \frac{60}{x^2} \, dx \), note that \( \frac{60}{x^2} = 60x^{-2} \). Find the antiderivative: \[ \int \frac{60}{x^2} \, dx = \int 60x^{-2} \, dx = 60 \left( -x^{-1} \right) = -\frac{60}{x}. \] Evaluate it from 1 to 3: \[ \left[-\frac{60}{x}\right]_1^3 = \left(-\frac{60}{3}\right) - \left(-\frac{60}{1}\right) = -20 + 60 = 40. \]

Step 4: Computing the Average Value

Substitute the result of the integral back into the equation for average value: \[ f_{avg} = \frac{1}{2} \times 40 = 20. \]

Step 5: Final Result

The average value of the function \( f(x) = \frac{60}{x^2} \) over the interval \( [1, 3] \) is 20.

Key Concepts

Definite Integral

The concept of a definite integral is at the heart of determining the average value of a function over a specific interval. A definite integral takes a function and computes the "net area" under the curve of the graph of that function, between two endpoints on the x-axis. This is represented by the notation \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the endpoints of the interval. In our original exercise, \( a = 1 \) and \( b = 3 \).

• The definite integral sums up infinitely small vertical slices of the area under the curve.
• The result gives the total accumulation of the function's values, taking changes and negative areas into account.

Understanding definite integrals can help you grasp how functions behave over intervals and pave the way to compute values like the average, as seen in the original problem.

Antiderivative

Finding the antiderivative is a crucial step in solving a definite integral. The antiderivative of a function is another function whose derivative is the original function. This is because integration, especially with indefinite integrals, is essentially the reverse process of differentiation.For example, consider the function \( f(x) = \frac{60}{x^2} = 60x^{-2} \). To find its antiderivative, we reverse the process of differentiation by applying the power rule in reverse:

• Take \( 60x^{-2} \).
• Increase the power by 1 to get \( 60x^{-1} \), and then divide by the new power.
• The antiderivative becomes \( -\frac{60}{x} \), which means the derivative of this function brings us back to our original \( 60x^{-2} \).

Knowing how to find antiderivatives is fundamental in evaluating definite integrals and hence computing the average value in our exercise.

Integration

Integration is the process by which we calculate the area under a curve for a given function. There are two types of integrals: definite and indefinite. In the context of the original exercise, integration refers primarily to the calculation of the definite integral, which ultimately helps us find the average value of the function.During integration, especially when working with power functions like \( f(x) = 60x^{-2} \), knowing rules like the reverse power rule simplifies the process. To integrate, you:

• Adjust the power, making it one higher.
• Normalize terms by dividing by this new power if integrating a standard power function.

The limits of integration, \( 1 \) and \( 3 \) in this case, are then used to evaluate the antiderivative at these points, showcasing how integration ties into the larger calculation process.

Interval

The interval defines the domain over which we evaluate the average value of a function. It is crucial because it sets the limits of our definite integral. In our exercise, the interval \( [1, 3] \) means we examine the behavior of the function starting from \( x = 1 \) to \( x = 3 \).

• An interval indicates where the integration begins and ends.
• It can affect the value of the definite integral significantly, as different intervals can capture different sections and curvatures of a graph.

For average value calculations, this chosen interval is divided into small sections as the definite integral is computed, and the sum over the entire interval is divided by the length of the interval \( b-a \). Therefore, understanding how to choose and utilize intervals is key to unlocking the correct results.