Problem 10 - Calculus Single Variable

Question

A function \(f\) is defined on a specified interval \(I=[a, b] .\) Calculate the area of the region that lies between the vertical lines \(x=a\) and \(x=b\) and between the graph of \(f\) and the \(x\) -axis. $$ f(x)=\left(x^{2}-1\right)(x-2) \quad I=[-2,2] $$

Step-by-Step Solution

Answer

The area between the curve and the x-axis on \([-2, 2]\) is 0.

Step 1: Determine the Integral Setup

To find the area between the curve and the x-axis from \(x = a\) to \(x = b\), we need to evaluate the definite integral \( \int_{a}^{b} f(x) \, dx \). For the given function \( f(x) = (x^2 - 1)(x - 2) \), the integral becomes \( \int_{-2}^{2} (x^2 - 1)(x - 2) \, dx \).

Step 2: Expand the Function

Expand the function \( f(x) = (x^2 - 1)(x - 2) \). First, multiply \(x^2 - 1\) by \(x\) to get \(x^3 - x\), then multiply \(-2\) with \(x^2 - 1\) to get \(-2x^2 + 2\). Combine them to get \( f(x) = x^3 - x - 2x^2 + 2 = x^3 - 2x^2 - x + 2 \).

Step 3: Integrate the Function

Integrate \( f(x) = x^3 - 2x^2 - x + 2 \) from \(-2\) to \(2\). The integral is: \[ \int_{-2}^{2} (x^3 - 2x^2 - x + 2) \, dx = \left[ \frac{x^4}{4} - \frac{2x^3}{3} - \frac{x^2}{2} + 2x \right]_{-2}^{2} \].

Step 4: Evaluate the Definite Integral

Substitute the limits \(-2\) and \(2\) into the integrated function:1. Evaluate at \(x = 2\): \(\frac{2^4}{4} - \frac{2 \times 2^3}{3} - \frac{2^2}{2} + 2 \times 2 = 4 - \frac{16}{3} - 2 + 4 = \frac{10}{3}\).2. Evaluate at \(x = -2\): \(\frac{(-2)^4}{4} - \frac{2 \times (-2)^3}{3} - \frac{(-2)^2}{2} + 2 \times (-2) = 4 + \frac{16}{3} - 2 - 4 = \frac{10}{3}\).3. Calculate the difference: \( \frac{10}{3} - \frac{10}{3} = 0 \).

Step 5: Interpret the Result

The result of the integral calculation, which is \(0\), implies that the area above the x-axis is completely balanced by the area below the x-axis within the given interval \([-2, 2]\). Therefore, the net area is zero.

Key Concepts

Calculating Area

When we talk about calculating the area under a curve, we're essentially trying to find out the size of the "space" that the function covers between two points on the x-axis. Imagine drawing a shape on graph paper and counting how many squares it covers. The concept of area under a curve is similar, but it's like trying to measure the space in a more fluid, continuous way.
To achieve this, we use calculus, specifically integration. In mathematical terms, the area under a curve from a point \(a\) to a point \(b\) is found using the definite integral of the function. Here's a simplified breakdown:

• First, identify the interval \([a, b]\) over which you want to measure the area.
• Set up an integral with these limits; \( \int_{a}^{b} f(x) \, dx \).
• Evaluate this integral to find the area.

This process can involve both positive and negative contributions if the curve dips below the x-axis. If these balance out perfectly, like in our example, the net area can be zero.

Polynomial Functions

Polynomial functions are mathematical expressions consisting of variables raised to whole number exponents, and they look like a sequence of terms added or subtracted together. For example, \( f(x) = (x^2 - 1)(x - 2) \) is a polynomial function.
Understanding polynomials involves looking at their structure:

• Terms: Individual parts of a polynomial, like \(x^2\) or \(-1\).
• Coefficients: Numbers multiplying the variable, such as \(-2\) in \(-2x\).
• Degree: The highest power the variable is raised to; here it is 3, because of \(x^3\).

These functions can have different shapes on a graph, like curves, peaks or troughs. By expanding the function as shown, we make it easier to integrate and find specific values. The structure gives it flexibility and the ability to describe complex shapes in a straightforward mathematical language.

Definite Integral Calculation

Definite integral calculation is a core concept in calculus used to quantify the area under a curve between two points. In our specific example, the process went as follows:
First, we identified our function as \(f(x) = x^3 - 2x^2 - x + 2\) and needed to calculate the integral over \([-2, 2]\). The steps to evaluate a definite integral include:

• Integrate: Find the antiderivative. For each term, this involves slightly increasing the exponent and dividing by the new exponent, transforming expressions like \(x^3\) into \(\frac{x^4}{4}\).
• Evaluate Boundaries: Substitute the upper and lower limits into the antiderivative and find the difference.
• Finding the Area: Calculate the result by subtracting the lower limit evaluation from the upper limit.

In this instance, the area turned out to be zero. This means that areas of regions above and below the x-axis within the specified interval exactly cancel each other out. The rich detail within definite integral calculations enables us to handle complex questions about areas and other quantities within various mathematical contexts.