Problem 39 - Calculus and its applications

Question

Find the average function value over the given interval. $$ f(x)=x^{2}+x-2 ; \quad[0,4] $$

Step-by-Step Solution

Answer

The average value of the function over the interval \([0, 4]\) is \(\frac{16}{3}\).

Step 1: Understand the Problem

To find the average value of a function over a specific interval, we use the formula for the average value of a function, which is: \[ f_{avg} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \] where \(a\) and \(b\) are the endpoints of the interval.

Step 2: Substitute in Values

In this problem, the interval is \([0, 4]\) and the function is \(f(x) = x^2 + x - 2\). Substitute \(a = 0\) and \(b = 4\) into the formula for the average value.

Step 3: Set Up the Integral

Set up the integral \( \int_{0}^{4} (x^2 + x - 2) \, dx \). This involves finding the antiderivative of \(f(x) = x^2 + x - 2\).

Step 4: Find the Antiderivative

The antiderivative of \(f(x) = x^2 + x - 2\) is \(F(x) = \frac{x^3}{3} + \frac{x^2}{2} - 2x + C\).

Step 5: Evaluate the Definite Integral

Evaluate the definite integral using the Fundamental Theorem of Calculus: \[ \int_{0}^{4} (x^2 + x - 2) \, dx = \left[ \frac{x^3}{3} + \frac{x^2}{2} - 2x \right]_{0}^{4} \] First, substitute \(x=4\): \[ \frac{4^3}{3} + \frac{4^2}{2} - 2(4) \] Then substitute \(x=0\): \[ \frac{0^3}{3} + \frac{0^2}{2} - 2(0) \].

Step 6: Calculate the Values

Calculate: For \(x=4\): \[ \frac{64}{3} + 8 - 8 = \frac{64}{3} \] For \(x=0\): \[ 0 \] So the integral evaluates to \[ \frac{64}{3} - 0 = \frac{64}{3} \].

Step 7: Find Average Value

The average value of the function is given by \[ f_{avg} = \frac{1}{4-0} \times \frac{64}{3} = \frac{64}{12} = \frac{16}{3} \].

Key Concepts

Definite Integral

A definite integral plays a crucial role in finding the area under a curve for a given interval. It's often used to calculate the exact value of this area using the function within integration bounds, defined by the limits. In the exercise, we are asked to compute the definite integral of the function \[f(x) = x^2 + x - 2\] over the interval \([0, 4]\). The definite integral is expressed as: \[\int_{0}^{4} (x^2 + x - 2) \, dx\] The process involves:

• Finding the antiderivative of the function.
• Applying the limits of integration, substituting each into the antiderivative, and subtracting the results.

Once the definite integral is evaluated, it provides the total area above the x-axis and below the curve over the interval specified. This concept forms the backbone of determining average values and accumulating quantities.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the world of differentiation and integration, showing how these two rich concepts are interconnected. It comprises two main parts, with the focus here often being on the second part, which connects definite integrals to antiderivatives. This theorem tells us that if we find an antiderivative \(F(x)\) of \(f(x)\), then:\[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\]In the exercise, this principle is used to evaluate the definite integral of \(f(x) = x^2 + x - 2\). We first find its antiderivative, which serves as the function \(F(x)\). Evaluating \(F(x)\) at the upper limit \(b = 4\) and the lower limit \(a = 0\), and subtracting them gives us:\[F(4) - F(0)\]This evaluation provides the net area under the curve, crucial for finding averages and totals in integral calculus problems.

Antiderivative

Antiderivatives, also known as indefinite integrals, are functions whose derivative results in the original function. Simply put, if you differentiate an antiderivative \(F(x)\), you'll return to your starting function \(f(x)\). In the given exercise, the function is:\[f(x) = x^2 + x - 2\]Finding the antiderivative means reversing the process of differentiation. The antiderivative can be expressed as:\[F(x) = \frac{x^3}{3} + \frac{x^2}{2} - 2x + C\]Where \(C\) represents the constant of integration, which is irrelevant in definite integrals, as it's canceled out during evaluation. This antiderivative is crucial because it allows us to apply the Fundamental Theorem of Calculus to calculate definite integrals, helping to find precise areas, averages, or total values over specified intervals.