Problem 46 - Applied Calculus

Question

Find the area bounded by the given curves. \(y=x^{2}-4\) and \(y=8-2 x^{2}\)

Step-by-Step Solution

Answer

The area bounded by the curves is 32 square units.

Step 1: Find the Points of Intersection

To find the region bounded by the curves, we first determine the points of intersection. Set the equations equal to each other: \[ x^2 - 4 = 8 - 2x^2 \]Rearrange and combine like terms:\[ 3x^2 = 12 \]Divide by 3:\[ x^2 = 4 \]Solving for \( x \), we get:\[ x = 2 \quad \text{or} \quad x = -2 \]Hence, the points of intersection are at \( x = -2 \) and \( x = 2 \).

Step 2: Determine the Intersection Points' Y-Coordinates

Substitute \( x = -2 \) and \( x = 2 \) back into either equation to find the y-coordinates.For \( x = -2 \):\[ y = (-2)^2 - 4 = 4 - 4 = 0 \]For \( x = 2 \):\[ y = (2)^2 - 4 = 4 - 4 = 0 \]So, the intersection points are \((-2, 0)\) and \((2, 0)\).

Step 3: Set Up the Integral for the Area

The area between the curves is obtained by integrating the difference between the top function and the bottom function from the left to the right intersection points.The top function is \( y = 8 - 2x^2 \) and the bottom function is \( y = x^2 - 4 \).The integral setup is:\[ \int_{-2}^{2} [(8 - 2x^2) - (x^2 - 4)] \, dx \]

Step 4: Simplify the Integral Expression

Simplify the expression inside the integral:\[ (8 - 2x^2) - (x^2 - 4) = 8 - 2x^2 - x^2 + 4 = 12 - 3x^2 \]Thus, the integral simplifies to:\[ \int_{-2}^{2} (12 - 3x^2) \, dx \]

Step 5: Calculate the Definite Integral

Now, compute the integral:\[ \int (12 - 3x^2) \, dx = 12x - x^3 \]Evaluate this from \(-2\) to \(2\):\[ \left[ 12x - x^3 \right]_{-2}^{2} \]Calculate at the upper limit:\[ 12(2) - (2)^3 = 24 - 8 = 16 \]Calculate at the lower limit:\[ 12(-2) - (-2)^3 = -24 + 8 = -16 \]Find the difference:\[ 16 - (-16) = 16 + 16 = 32 \]

Step 6: Conclude the Area Calculation

The total area bounded by the curves \( y = x^2 - 4 \) and \( y = 8 - 2x^2 \) is 32 square units.

Key Concepts

Integral Calculus

Integral Calculus is a branch of mathematics that focuses on integrals and their properties. Understanding this concept is essential for finding areas under curves, among many other applications. There are two main types of integrals in calculus: indefinite and definite integrals.

• Indefinite Integrals: These refer to antiderivatives and do not have specified limits. They represent a family of functions.
• Definite Integrals: These have upper and lower limits and are used to calculate the net area under a curve between these limits.

In the given problem, integral calculus helps to determine the area between two curves, which involves both definite integrals and the concept of integration as an area accumulation. By setting up the proper integral bounds, we're able to find the precise area that lies between the two mathematical curves.

Definite Integral

A Definite Integral is an integral that evaluates to a number, offering a quantitative measure of what is described by the integral's function over a specified interval. This concept is crucial in calculating bounded areas, as it enables us to find how much space a function covers within a certain range.
In the example problem, we use the definite integral to find the area between the curves defined by the equations:

• \(y = x^2 - 4\)
• \(y = 8 - 2x^2\)

The definite integral is set up between the points of intersection: for this specific problem, from \(x = -2\) to \(x = 2\). By simplifying the integrand, we confirmed the function we needed to integrate and evaluate over these bounds. This approach results in assessing the exact area contained between these intersecting curves.

Points of Intersection

The Points of Intersection of two curves occur where the graphs of the equations meet. Finding these points is an important step when calculating the area between curves, as they establish the integration limits.
To find these points in the example, we set the equations equal to each other: \(x^2 - 4 = 8 - 2x^2\).

• First, simplify and solve the equation for \(x\). The solution gives \(x = 2\) and \(x = -2\).

These values are confirmed by plugging them back into either equation, both of which resulted in \(y = 0\). Thus, the intersection points, (-2, 0) and (2, 0), define the interval over which we calculate the area using integration. Recognizing and correctly identifying these points is integral to ensuring accurate area computation.