Question

Find the average value of the function on the given interval. $$ f(x)=4 x^{3} ; \quad[1,3] $$

Short answer

The average value is 40.

Step-by-step solution

Understand the Formula for Average Value of a Function

The average value of a function \( f(x) \) on the interval \( [a, b] \) is given by the formula: \[ \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]. Identify the function and the interval from the problem. Here \( f(x) = 4x^3 \) and the interval \([1, 3]\).

Set Up the Integral

Substitute the function \( f(x) = 4x^3 \) and the interval \([1,3]\) into the formula for the average value. This gives: \[ \frac{1}{3-1} \int_{1}^{3} 4x^3 \, dx \]. Simplify to: \[ \frac{1}{2} \int_{1}^{3} 4x^3 \, dx \].

Calculate the Integral

Integrate the function \( 4x^3 \) over the interval \([1, 3]\): \[ \int 4x^3 \, dx = x^4 + C \]. Evaluate the definite integral: \[ \int_{1}^{3} 4x^3 \, dx = \left[ x^4 \right]_1^3 = 3^4 - 1^4 = 81 - 1 = 80 \].

Divide by the Length of the Interval

Now divide the result of the integral, 80, by the length of the interval, which is 2: \[ \frac{1}{2} \times 80 = 40 \].

State the Average Value

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

Key concepts

Definite Integral

The concept of the definite integral is pivotal in understanding how functions behave over intervals. In simple terms, a definite integral helps to find the area under the curve of a function within a specified range. For a function \(f(x)\), evaluated over an interval \([a, b]\), the integral is represented as \[ \int_{a}^{b} f(x) \, dx \].
This notation essentially means adding up countless infinitesimally small products of \(f(x)\) values and their corresponding tiny changes in \(x\), across the interval from \(a\) to \(b\).

• The definite integral gives us the net area, which could be positive, negative, or zero, depending on the function's behavior and the chosen interval.
• Calculating a definite integral requires an understanding of antiderivatives, which are functions that, when differentiated, yield the original \(f(x)\).
• In practical problems, like finding average values, the definite integral is vital as it provides the accumulated value across a range, which can then be used to derive averages.

Integral Calculus

Integral calculus is one part of calculus, the other being differential calculus. While differential calculus focuses on the rate of change (derivatives), integral calculus deals with accumulation, like areas under curves.
In the context of integral calculus, we are often concerned with the antiderivative, sometimes called the primitive, of a function. Finding an antiderivative means determining a function \(F(x)\) such that its derivative \(F'(x)\) equals \(f(x)\).

• Two integral types are commonly studied: indefinite integrals (without limits) and definite integrals (with specific limits, like \([a, b]\)).
• In practice, these integrals allow us to solve a broad range of problems in physics, engineering, and economics, where understanding the total quantity accumulated over time or distance is crucial.
• For computing an average value of a function, as in the given exercise, integral calculus is essential as it provides the necessary methodology to evaluate and apply the integral over a specific range.

Calculating Integral

Calculating an integral involves a step-by-step process that requires understanding of both the function and the limits of integration. Here's how you do it:
1. **Setup the Integral**: Identify the function and the limits over which you need to integrate. The problem provides \(f(x) = 4x^3\) and the interval \([1, 3]\).
2. **Antidifferentiate the Function**: Determine the antiderivative of \(f(x)\). In this case, the antiderivative of \(4x^3\) is \(x^4 + C\), where \(C\) represents the constant of integration.
3. **Evaluate the Definite Integral**: Substitute the limits into the antiderivative. For our example, compute \[ \left[ x^4 \right]_1^3 = 3^4 - 1^4 = 81 - 1 = 80 \]. This is the value of the definite integral.
4. **Apply to the Context**: If you're calculating an average, divide the result by the interval's width. Since the length of the interval is 2, and the result of the integration is 80, the final average value is \[ \frac{1}{2} \times 80 = 40 \].
By following these steps, you'll correctly compute both the integral and any relevant average, such as in this example.