Problem 56 - Calculus: Early Transcendental Functions

Question

Find the average value of the function on the given interval. \(f(x)=x^{3}-3 x^{2}+2 x,[1,2]\)

Step-by-Step Solution

Answer

The average value of the function on the given interval is \(\frac{1}{2}\).

Step 1: Formulate the Setup

Given the function \(f(x)=x^{3}-3 x^{2}+2 x\), on the interval [1,2]. To find the average, the formula \(\frac{1}{b-a}\int_{a}^{b} f(x) dx\) will be used. Here, \(a=1\) and \(b=2\).

Step 2: Find Integral of Function

The integral of \(f(x)\) is calculated as: \(\int f(x) dx = \int (x^{3} - 3x^{2} + 2x) dx = \frac{1}{4}x^{4} - x^{3} + x^{2}\)

Step 3: Plug in Limit Values and Calculate

Now, substitute the values of \(a\) and \(b\) into the integral of the function, then subtract integral at \(a\) from integral at \(b\): \(\int_{1}^{2} f(x) dx = [\frac{1}{4}*2^{4} - 2^{3} + 2^{2}] - [\frac{1}{4}*1^{4} - 1^{3} + 1^{2}] = \frac{1}{2}\)

Step 4: Calculate Average Value of Function

Finally, substitute the calculated integral and the limits into the formula for the average value of a function to get the final solution. Average Value = \(\frac{1}{2-1}\int_{1}^{2} f(x) dx = \frac{1}{1}*\frac{1}{2} = \frac{1}{2}\).

Key Concepts

Definite Integral

The concept of a definite integral is key when measuring the total "accumulation" of a quantity. It gives us the precise value of the area under the curve of a function between two points on the x-axis. In our example, we used the definite integral to find the average value of the function

• The definite integral is calculated between set boundaries, indicated by the limits at the top and bottom of the integral sign.
• For the function \(f(x) = x^3 - 3x^2 + 2x\) on [1, 2], the definite integral will explore how the function behaves between these x-values.

This approach not only shows the total change in the function over the interval but also is essential for solving problems related to the area between the curve and the x-axis. It's a foundational concept in calculus and helps in various real-world applications such as physics, engineering, and economics.

Integration Techniques

Integration techniques are crucial tools used to find integrals of various functions. Sometimes, functions can be simple, and their integrals can be found using basic rules. Other times, more sophisticated methods are required. The choice of technique often depends on the form of the function you're working with.

• In the example, our function, \( x^3 - 3x^2 + 2x \), is a polynomial, so we applied the power rule of integration.
• The power rule states that to integrate \( x^n \), you add 1 to the exponent and divide by the new exponent: \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \).

This simple method keeps the finding of the antiderivative straightforward. Beyond polynomials, there are techniques like integration by parts and trigonometric substitutions designed for more complex integrals. By understanding when and how to use each technique, you can solve intricate calculus problems more effectively.

Polynomial Functions

Polynomial functions are expressed in the form of \(a_nx^n + a_{n-1}x^{n-1} + \, ... \, + a_1x + a_0\). These are functions that are composed of variables and coefficients, using only the operations of addition, subtraction, multiplication, and non-negative integer exponents. In many calculus problems, like ours, polynomial functions are common due to their predictability and ease of manipulation.

• A clear characteristic of polynomials is their smooth and continuous nature, which makes them easy to integrate.
• In our given function, \(f(x) = x^3 - 3x^2 + 2x\), each term is a product of a constant and a power of \(x\), showing it is a third-degree polynomial.
• The structure of polynomial functions allows for straightforward application of the power rule when integrating, as shown in our solution process.

Their consistent behavior and the mathematical principles they share make them a favorite in mathematical modeling and solving practical problems.