Problem 36 - Calculus: Early Transcendental Functions

Question

Compute the average value of the function on the given interval. $$f(x)=x^{2}+2 x,[0,1]$$

Step-by-Step Solution

Answer

The average value of the function \(f(x)=x^{2}+2 x\) on the interval \[0,1]\ is \(\frac{4}{3}\).

Step 1: Identify the interval and the function

For \(f(x)=x^{2}+2 x\), the interval is from a=0 to b=1.

Step 2: Insert the values into the formula

The formula to find the average value of a function is \[ \frac{1}{b - a} \int_{a}^{b} f(x) dx \]. Insert 'a' as 0, 'b' as 1 and \(f(x)\) as \(x^{2}+2 x\) into the formula: \[ \frac{1}{1 - 0} \int_{0}^{1} (x^{2}+2 x) dx \]

Step 3: Evaluate the integral

Now, evaluate the integral \[ \int_{0}^{1} (x^{2}+2 x) dx \]. The antiderivative of \(x^{2}+2 x\) is \[\frac{x^{3}}{3} + x^{2}\] and evaluating it from 0 to 1 gives: \[ \[\left(\frac{1^{3}}{3} + 1^{2}\right) - \left(\frac{0^{3}}{3} + 0^{2}\right) = \frac{1}{3} + 1 = \frac{4}{3} \]

Step 4: Compute the average value of the function

Put the value of the integral into the formula to find the average value: \[\frac{1}{1} \cdot \frac{4}{3} = \frac{4}{3}\].

Key Concepts

Definite Integrals

Understanding definite integrals is vital when calculating the average value of a function over a given interval. Essentially, a definite integral takes a function and computes the net area under the curve between two specific points on the x-axis, denoted as the limits of integration.

In our exercise, the definite integral is used with the lower limit 'a' as 0 and the upper limit 'b' as 1 for the function f(x) = x^2 + 2x. This process involves finding the limit of a sum as the number of rectangles under the curve increases and the width of these rectangles approaches zero. The result of this calculation tells us the accumulated value of our function between the specified limits of 0 and 1.

When you're solving problems involving the average value of a function, you'll often set up a definite integral like this one to start. It's important to note that the definite integral gives you a numerical value, indicating the area or accumulated quantity, which will be key in computing the average.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a significant principle that bridges the concept of an antiderivative with that of a definite integral. It is composed of two parts, with the first stating that the integral of a function over an interval can be found by evaluating its antiderivative at the endpoints.

In the context of our exercise, when we have the integral of f(x) = x^2 + 2x from 0 to 1, the Fundamental Theorem of Calculus allows us to compute the integral by finding the antiderivative and then subtracting its value at the lower limit from its value at the upper limit. This process turns a potentially complex area calculation into a simple substitution, radically simplifying the computational work necessary.

Understanding this theorem is crucial because it not only simplifies the calculation of definite integrals but also highlights a deep connection between differential and integral calculus, reinforcing how the concepts of differentiation and integration are inverse processes.

Antiderivatives

Antiderivatives, or indefinite integrals, are the counterpart to derivatives in calculus. An antiderivative of a function is another function that, when differentiated, yields the original function. It represents the collection of all the functions that share the same derivative, differing only by a constant.

For our function f(x) = x^2 + 2x, the antiderivatives are F(x) = (x^3/3) + x^2 + C, where C represents the constant of integration. During the solution process, this constant doesn't affect the calculation since it cancels out when evaluating a definite integral - a result of the Fundamental Theorem of Calculus.

To find the average value of a function, you first need to identify its antiderivative, apply the limits of integration, and then divide by the length of the interval. This process of integration and averaging involves working backwards from differentiation, illustrating the importance of having a solid grasp on the concept of antiderivatives.