Problem 34 - Calculus and its applications

Question

Find the average function value over the given interval. $$ y=2 x^{3} ; \quad[-1,1] $$

Step-by-Step Solution

Answer

The average value of the function \( y = 2x^3 \) over the interval \([-1, 1]\) is 0.

Step 1: Set up the average value formula

The average value of a function \( f(x) \) over the interval \([a, b]\) is given by the formula: \[ f_{avg} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx \]In this exercise, \( f(x) = 2x^3 \), and the interval \([a, b]\) is \([-1, 1]\). Substitute these values into the formula: \[ f_{avg} = \frac{1}{1 - (-1)} \int_{-1}^{1} 2x^3 \,dx \]

Step 2: Simplify the expression

Calculate the width of the interval \( b - a = 1 - (-1) = 2 \). This modifies the expression to: \[ f_{avg} = \frac{1}{2} \int_{-1}^{1} 2x^3 \,dx \]

Step 3: Evaluate the integral

Find the indefinite integral of \( 2x^3 \):\[ \int 2x^3 \, dx = \frac{2}{4}x^4 + C = \frac{1}{2}x^4 + C \]Now evaluate the definite integral from \(-1\) to \(1\): \[ \int_{-1}^{1} 2x^3 \, dx = \left[ \frac{1}{2}x^4 \right]_{-1}^{1} = \left( \frac{1}{2}(1)^4 \right) - \left( \frac{1}{2}(-1)^4 \right) \]

Step 4: Calculate the definite integral

Perform the calculations for the definite integral:\[ \frac{1}{2}(1)^4 = \frac{1}{2} \]\[ \frac{1}{2}(-1)^4 = \frac{1}{2} \]Subtract the two results:\[ \frac{1}{2} - \frac{1}{2} = 0 \]

Step 5: Compute the average value

Substitute the result of the definite integral back into the formula for the average value:\[ f_{avg} = \frac{1}{2} \times 0 = 0 \] Hence, the average value of the function \( y = 2x^3 \) over the interval \([-1, 1]\) is 0.

Key Concepts

Definite Integral

The concept of the definite integral is fundamental in calculus. It helps measure the total amount accumulated by a function over a specific interval. For a function \( f(x) \), the definite integral is written as \( \int_{a}^{b} f(x) \, dx \), where \([a, b]\) is the interval of interest. This expression essentially calculates the "net area" between the function \( f(x) \) and the x-axis, from x = a to x = b.

• The definite integral gives a precise number that can represent various physical quantities like area, volume, or even net change of a phenomenon.
• In terms of the exercise, we use the definite integral to find how much the function \( y = 2x^3 \) sums up over the interval \([-1, 1]\).
• To solve it, we need to first find the antiderivative or the indefinite integral and then evaluate it at the interval limits.

Indefinite Integral

An indefinite integral, also known as an antiderivative, is the reverse process of taking a derivative. When we find an indefinite integral of a function, we are basically asking: "What function did we originally take the derivative of to get this?"The indefinite integral is represented as \( \int f(x) \, dx \), and it includes a constant of integration \( C \).Let's break it down:

• The indefinite integral of \( 2x^3 \) is \( \int 2x^3 \, dx = \frac{1}{2}x^4 + C \). This shows the family of functions whose derivative is \( 2x^3 \).
• Finding this antiderivative is the first step in calculating definite integrals as well! As you can see, calculating both indefinite and definite integrals are tied together.

In the context of the exercise, this process is necessary to evaluate the definite integral from -1 to 1 to ultimately find the average value of the function.

Integral Calculus

Integral calculus is an essential branch of calculus focused on integration and its processes. It serves to find functions from their derivatives and to compute things like areas under curves or accumulations of quantities. Here are some key points about integral calculus:

• Integration: The main focus of integral calculus, where we reverse the process of differentiation. Two primary types are indefinite and definite integrals.
• Applications: Used in physics, engineering, economics, and beyond, integral calculus helps calculate distances, areas, volumes, and other accumulations. It shows up in practical problems like finding the average value of functions.
• The Fundamental Theorem of Calculus: This important principle links differentiation and integration, stating that differentiation and integration are inverse processes.

In this exercise, integral calculus is applied to find the average value of a function on a specific interval using both definite and indefinite integrals.