Problem 58 - University Calculus: Early Transcendentals

Question

Find the total area between the region and the \(x\) -axis. $$y=3 x^{2}-3, \quad-2 \leq x \leq 2$$

Step-by-Step Solution

Answer

The total area between the region and the x-axis is 10 square units.

Step 1: Identify the Function and Interval

The function given is \( y = 3x^2 - 3 \) and the interval is from \( x = -2 \) to \( x = 2 \). We need to find the area between this curve and the x-axis over this interval.

Step 2: Find the Points of Intersection with the x-axis

To find where the function intersects the x-axis, set \( y = 0 \): \[ 3x^2 - 3 = 0 \]Solving for \( x \), we get \( x^2 = 1 \), which gives roots \( x = 1 \) and \( x = -1 \). The function intersects the x-axis at these points.

Step 3: Determine Area Slices Using the x-axis Intersections

We now consider the sub-intervals defined by the points of intersection: \([-2, -1]\), \([-1, 1]\), and \([1, 2]\). On these sub-intervals, calculate the area between the curve and the x-axis. Additionally, determine whether to use the absolute value of the integral to account for below-x-axis regions.

Step 4: Calculate the Integral for Each Sub-interval

Calculate the definite integral of \( y \) from \( x = -2 \) to \( x = -1 \):\[ \int_{-2}^{-1} (3x^2 - 3) \, dx \] Calculate the definite integral of \( y \) from \( x = -1 \) to \( x = 1 \): \[ \int_{-1}^{1} (3x^2 - 3) \, dx \] Finally, calculate the definite integral of \( y \) from \( x = 1 \) to \( x = 2 \): \[ \int_{1}^{2} (3x^2 - 3) \, dx \]

Step 5: Evaluate Each Integral

Evaluate each integral separately:1. For \( \int_{-2}^{-1} (3x^2 - 3) \, dx \): \[ \int_{-2}^{-1} (3x^2 - 3) \,dx = \left[ x^3 - 3x \right]_{-2}^{-1} = (-1 - 3(-1)) - (-8 + 6) = 2 \]2. For \( \int_{-1}^{1} (3x^2 - 3) \, dx \): \[ \int_{-1}^{1} (3x^2 - 3) \, dx = \left[ x^3 - 3x \right]_{-1}^{1} = (1 - 3) - (-1 + 3) = -4 \]3. For \( \int_{1}^{2} (3x^2 - 3) \, dx \): \[ \int_{1}^{2} (3x^2 - 3) \, dx = \left[ x^3 - 3x \right]_{1}^{2} = (8 - 6) - (1 - 3) = 4 \]

Step 6: Calculate Total Area Using Absolute Values

For the total area between the curve and the x-axis, sum the absolute values from each integral (since area can't be negative):- Absolute Area from \([-2, -1]\) is \(|2| = 2\).- Absolute Area from \([-1, 1]\) is \(|-4| = 4\).- Absolute Area from \([1, 2]\) is \(|4| = 4\).Total area is \(2 + 4 + 4 = 10\).

Key Concepts

Area Under Curve

The area under a curve refers to the total space located between the graph of a function and the x-axis over a specified interval. In mathematical terms, this is represented by the definite integral of the function over that interval. Calculating this area is crucial in many applications, from physics to finance.
For the given function, the task is to compute the total area enclosed by the curve of the polynomial function and the x-axis from the interval [-2, 2].
It's essential to determine whether the curve is above or below the x-axis, as negative values affect the calculation of area. Thus, while evaluating the definite integral, taking the absolute value when necessary ensures the correct measure of the area without considering negativity.

Polynomial Functions

A polynomial function is an expression involving variables raised to whole number powers and multiplied by coefficients. Understanding polynomial functions is vital for analyzing different types of graphs and their properties
In this scenario, we have the polynomial function:
\[ y = 3x^2 - 3 \].
This is a quadratic polynomial, characterized by the highest degree term being \(x^2\). This kind of polynomial typically forms a parabolic curve when graphed.

• The leading term \(3x^2\) suggests the parabola opens upwards.
• Knowing the general symmetry and shape of quadratic polynomials can assist with determining critical parts of the graph relevant to area calculations like vertex, axis of symmetry, and intersection points.

Definite Integral Calculation

The definite integral is a powerful tool in calculus used to compute the net area between the curve represented by a function and the x-axis. Specifically, the definite integral evaluates how much the function accumulates or depletes over a defined interval.
For our exercise, we need to compute the definite integrals of the function \( y = 3x^2 - 3 \) over three separate sub-intervals :-

• \([-2, -1]\)
• \([-1, 1]\)
• \([1, 2]\)

Evaluating each part ensures we correctly manage transitions from being above or below the x-axis. This approach helps correctly compute and combine the total area.
For each interval, integrate and evaluate: subtract the value of the integral at the starting point from the value at the endpoint. Sum the absolute values of these integrals to get the total area.

Intersection Points with x-axis

Finding where the polynomial function intersects the x-axis is crucial for determining segments that require separate integral evaluations. Intersection points are where the function's value \( y = 0 \), meaning the graph touches or crosses the horizon.
To find these, solve for \( x \) when \( y = 0 \):
\[ 3x^2 - 3 = 0 \]
Simplifying gives \( x^2 = 1 \), leading to roots \( x = 1 \) and \( x = -1 \).

• These points divide our interval into sections: some parts of the curve will lie above the x-axis, and others below.
• Each segment requires separate assessment via definite integration to assure the entire area is accounted for correctly.

Understanding these intersections supports accurate area computation by clarifying which intervals to integrate separately and whether a region is positive or negative relative to the x-axis.