Problem 11 - 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 is \(-\frac{2}{3}\).

Step 1: Understand the Average Value Formula for a Function

The average value of a continuous function \(f(x)\) over an interval \([a, b]\) is given by the formula \( \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \). In this case, \(f(x) = x^2 - 2\), \(a = 0\), and \(b = 2\).

Step 2: Set Up the Integral

Substitute the given function and interval limits into the formula. The expression becomes \( \frac{1}{2-0} \int_{0}^{2} (x^2 - 2) \, dx \). Simplifying, it becomes \( \frac{1}{2} \int_{0}^{2} (x^2 - 2) \, dx \).

Step 3: Compute the Definite Integral

Calculate the definite integral \(\int_{0}^{2} (x^2 - 2) \, dx\). The integral of \(x^2\) is \(\frac{x^3}{3}\) and the integral of \(-2\) is \(-2x\). Therefore, the expression becomes \( \left[ \frac{x^3}{3} - 2x \right]_{0}^{2} \).

Step 4: Evaluate the Integral at the Bounds

Plug in the upper bound 2 into the antiderivative: \(\frac{2^3}{3} - 2(2) = \frac{8}{3} - 4\). Then plug in the lower bound 0: \(\frac{0^3}{3} - 2(0) = 0\). Subtract the lower bound result from the upper bound result: \( \left( \frac{8}{3} - 4 \right) - 0 = \frac{8}{3} - 4 = \frac{8}{3} - \frac{12}{3} = -\frac{4}{3} \).

Step 5: Find the Average Value

Multiply the result of the integral by \(\frac{1}{2}\): \(\frac{1}{2} \times -\frac{4}{3} = -\frac{4}{6} = -\frac{2}{3}\). This is the average value of the function \(f(x)=x^2-2\) over the interval \([0,2]\).

Key Concepts

Definite Integral

The concept of a definite integral is a fundamental building block in calculus. It provides a way to calculate the accumulation of quantities, such as areas under curves, over a given interval. The definite integral of a function \( f(x) \) from \( a \) to \( b \) is written as \( \int_{a}^{b} f(x) \, dx \). This notation signifies the total accumulation of the function's values over the specified interval.Definite integrals have many practical applications:

• Area under a curve: You can determine the area between the curve of the function and the x-axis within the interval \([a, b]\).
• Physical quantities: It can represent the total distance traveled by an object when given its speed-time graph.
• Average values: In this context, the definite integral helps find the average value of a function over an interval, by averaging its accumulated value.

To find the definite integral, follow steps like setting up the integral with the given limits, computing the antiderivative of the function, and then evaluating it at the upper and lower bounds of the interval.

Continuous Function

A function is considered continuous if, when you draw it, there are no breaks, jumps, or holes in the graph. In mathematical terms, a function \( f(x) \) is continuous over an interval if it doesn't "break" at any point within that interval. Simply put, the function's value changes smoothly and gradually without sudden changes or gaps.Characteristics of a continuous function include:

• No jumps: The graph doesn’t have abrupt vertical shifts.
• No breaks: You can draw the graph without lifting your pencil.
• Smoothness: There are no sudden changes in direction.

Continuous functions are important in calculus for several reasons:

• Integrability: You can always find a definite integral of a continuous function over a closed interval.
• Predictability: Predictions based on the continuous function tend to be more reliable due to the absence of abrupt changes.

Understanding whether a function is continuous is crucial before calculating an integral because it guarantees that the integral (and hence the average value over an interval) is well-defined.

Interval

An interval in mathematics defines the scope within which you are analyzing or working with a function. It is essentially a segment of the real number line that contains all points between a starting point \( a \) and an ending point \( b \).Types of intervals include:

• Closed interval: Denoted as \([a, b]\), it includes both endpoints \( a \) and \( b \).
• Open interval: Written as \((a, b)\), it excludes the endpoints, considering only the points inside.
• Semi-open intervals: These can be \([a, b)\) or \((a, b]\), including only one endpoint.

In the provided exercise, the interval \([0, 2]\) means that we are finding the average value of the function between \( x = 0 \) and \( x = 2 \), including both 0 and 2. Understanding the type of interval is key when solving problems in calculus, as it determines the boundaries within which you calculate integrals or other mathematical properties.