Problem 25 - Calculus for Biology and Medicine

Question

Let \(f(x)=x^{2}-2\). Compute the average value of \(f(x)\) over the interval \([0,2]\).

Step-by-Step Solution

Answer

The average value of \( f(x) = x^2 - 2 \) over \([0, 2]\) is \(-\frac{2}{3}\).

Step 1: Recall the Formula for Average Value

The average value of a function \( f(x) \) on the interval \([a, b]\) is given by: \[ f_{avg} = \frac{1}{b-a} \int_a^b f(x) \, dx \].For this problem, \( a = 0 \) and \( b = 2 \), so the formula becomes \[ f_{avg} = \frac{1}{2-0} \int_0^2 (x^2 - 2) \, dx \].

Step 2: Set Up the Integral

Substitute the expression for \( f(x) \) into the integral:\[ \int_0^2 (x^2 - 2) \, dx \].

Step 3: Compute the Integral

Find the integral of \( x^2 - 2 \) within the limits from 0 to 2:\[ \int (x^2 - 2) \, dx = \int x^2 \, dx - \int 2 \, dx \].Compute each integral:\[ \int x^2 \, dx = \frac{x^3}{3} \quad \text{and} \quad \int 2 \, dx = 2x \].So, \[ \int_0^2 (x^2 - 2) \, dx = \left[ \frac{x^3}{3} - 2x \right]_0^2 \].

Step 4: Evaluate the Integral

Substitute the limits 0 and 2 into the result of the integral:\[ \left( \frac{2^3}{3} - 2(2) \right) - \left( \frac{0^3}{3} - 2(0) \right) \].Calculate the expression:\[ \frac{8}{3} - 4 \].Thus, the evaluated integral is \[ -\frac{4}{3} \].

Step 5: Calculate the Average Value

Substitute the result of the integral into the average value formula:\[ f_{avg} = \frac{1}{2} \left(-\frac{4}{3}\right) = -\frac{2}{3} \].

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics that deals with integrals and their properties. It involves two major operations: finding the **antiderivative** (often called the indefinite integral) and computing the **definite integral** of a function. Integrals allow us to compute areas under curves, total accumulated quantities, and many more applications.
When we talk about integrals, we often refer to the process of finding the area under a curve defined by a function over a specific interval. In essence, an integral can be thought of as the sum of an infinite number of infinitesimally small quantities.

• The symbol for integration is \( \int \), which is followed by a function and the variable of integration, such as \( \int f(x) \, dx \).
• Initially, focus on understanding the **geometric interpretation**: the area under the curve between two points on the x-axis.

Integral calculus is fundamental to many areas, including physics, engineering, and statistics, because it provides tools to model and analyze real-world situations.

Definite Integral

A definite integral is an integral that calculates the **exact net area** under a function between two specified limits, usually denoted by \( a \) and \( b \). It's called "definite" because it gives a specific numerical result, unlike an indefinite integral which produces a general form. The standard notation is \( \int_a^b f(x) \, dx \).
The definite integral has various properties that make it very useful:

• It is interpreted as the **net area** between the curve and the x-axis, considering areas above the x-axis as positive and those below as negative.
• The area calculation can be negative, meaning there is more area below the x-axis than above within the interval.
• If the function is continuous on the closed interval \[ a, b \], the **Fundamental Theorem of Calculus** helps link the definite integral to antiderivatives.

The exercise related to \( f(x) = x^2 - 2 \) over \[0, 2 \] involves calculating this definitive measure, resulting in the expression \( -\frac{4}{3} \). This tells us about the balance of the area above and below the x-axis.

Average Value Formula

The average value of a function over a certain interval gives the mean value of the function’s output across that interval. This is especially useful in real-world applications where you need to know not just the total or final value but the average behavior of a function over time or space.
The formula for the average value of a continuous function \( f(x) \) over an interval \[ a, b \] is: \[f_{avg} = \frac{1}{b-a} \int_a^b f(x) \, dx\]Here, \[- \( \frac{1}{b-a} \) acts as a normalizing factor, adjusting the integral to reflect the average over the interval's length.\] In the given problem, with \( f(x) = x^2 - 2 \) from \[0, 2 \], we substitute into the formula:

• First, calculate the integral \( \int_0^2 (x^2 - 2) \, dx \), which equals \( -\frac{4}{3} \).
• Then apply the average value formula: \( f_{avg} = \frac{1}{2} \left(-\frac{4}{3}\right) = -\frac{2}{3} \).

The result \( -\frac{2}{3} \) indicates that the average of this function's values is negative over the interval, suggesting a predominance of its values being below the x-axis.