Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-14, find the average value of the function on the given interval. 1\. \(f(x)=4 x^{3}, \quad[1,3]\)

Step-by-Step Solution

Answer

The average value is 40.

Step 1: Identify the Problem

We need to find the average value of the function \( f(x) = 4x^3 \) on the interval [1, 3].

Step 2: Recall the Formula for Average Value

The formula for the average value of a continuous function \( f(x) \) over the interval \( [a, b] \) is given by: \[ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \].

Step 3: Set Up the Integral

Substitute \( f(x) = 4x^3 \), \( a = 1 \), and \( b = 3 \) into the formula: \[ \text{Average value} = \frac{1}{3-1} \int_{1}^{3} 4x^3 \, dx \].

Step 4: Simplify the Integral Setup

This simplifies to: \[ \text{Average value} = \frac{1}{2} \int_{1}^{3} 4x^3 \, dx \].

Step 5: Evaluate the Integral

First, find the antiderivative of \( 4x^3 \), which is \( x^4 \). Now, evaluate the definite integral: \[ \int_{1}^{3} 4x^3 \, dx = \left[ x^4 \right]_{1}^{3} = 3^4 - 1^4 = 81 - 1 = 80 \].

Step 6: Calculate Average Value

Plug the value of the integral back into the average value formula: \[ \text{Average value} = \frac{1}{2} \times 80 = 40 \].

Step 7: Conclusion

The average value of the function \( f(x) = 4x^3 \) on the interval [1, 3] is 40.

Key Concepts

Integral Calculus

Integral calculus is all about understanding and working with integrals, which can be thought of as antiderivatives. Integrals help us find things like the area under curves or the total accumulation of quantities over an interval. There are two main types: **indefinite integrals** and **definite integrals**.

• **Indefinite integrals** represent a family of functions without specific limits, and they give us the general form of antiderivatives.
• **Definite integrals**, on the other hand, have specific bounds and give a concrete number representing the accumulated value across a particular interval.

In this exercise, finding the average value of a function involves calculating a definite integral, which provides the integration across the interval [1, 3]. Understanding integral calculus allows us to link the accumulated area to real-world applications, like finding average function values.

Continuous Function

A continuous function is one without any breaks, jumps, or holes in its graph. These functions are predictable and do not have any sudden changes in value.

• For a function to be continuous on an interval, it must be continuous at every point in that interval.
• Continuous functions have a great property: they can be integrated smoothly across any interval.

Our function, given as \(f(x) = 4x^3\), is a polynomial, making it continuous everywhere. This nature is crucial because only then can we apply the methods of integral calculus effectively to compute the required average value.

Definite Integral

The definite integral calculates the exact accumulation or total of a function's values over a closed interval \([a, b]\). It is represented using the integral sign with limits, like \(\int_{a}^{b} f(x) \, dx\).

• Here, the lower and upper bounds (\(a\) and \(b\)) are clear, defining where to start and stop accumulating.
• The process involves finding the antiderivative of \(f(x)\) and evaluating it at both bounds, then subtracting these results.

In the given solution, the definite integral \(\int_{1}^{3} 4x^3 \, dx\) was calculated to find the total area under the curve \(4x^3\) from \(x = 1\) to \(x = 3\). This definite integral contributes directly to determining the average value of the function on that interval.

Antiderivative

The antiderivative is essentially the reverse of taking a derivative. It is the function from which the original function could have been derived.

• Finding an antiderivative involves determining a function whose derivative is the given function.
• For example, the antiderivative of \(4x^3\) is \(x^4\), since the derivative of \(x^4\) gives us back \(4x^3\).

In our exercise, the antiderivative is crucial because it allows us to evaluate the definite integral over the desired interval. Once the antiderivative \(x^4\) was found, the next step was to compute it at the boundary values \(3\) and \(1\), leading to the calculation of the definite integral as part of finding the average value of the function.