Problem 21 - Basic Technical Mathematics with Calculus

Question

Find the areas bounded by the indicated curves. $$y=x^{4}-8 x^{2}+16, y=16-x^{4}$$

Step-by-Step Solution

Answer

The area between the curves is \(\frac{128}{15}\).

Step 1: Understand the Problem

We need to find the area between the curves given by the functions \(y = x^4 - 8x^2 + 16\) and \(y = 16 - x^4\). The area is bounded by these curves and lies between the points where the curves intersect.

Step 2: Find Points of Intersection

To find the points of intersection, set the equations equal: \(x^4 - 8x^2 + 16 = 16 - x^4\). Simplifying gives \(2x^4 - 8x^2 = 0\), which factors to \(2x^2(x^2 - 4) = 0\). Solving gives \(x = 0, x = 2, x = -2\).

Step 3: Set Up the Integral for Area

The area between two curves \(y_1\) and \(y_2\) (where \(y_1 > y_2\)) over the interval \([a, b]\) is given by \(\int_a^b (y_1 - y_2) \: dx\). Here, \(y_1 = 16 - x^4\) and \(y_2 = x^4 - 8x^2 + 16\).

Step 4: Identify Integration Limits and Functions

The points of intersection give the limits as \([-2, 2]\). Thus, set the integral: \[\int_{-2}^{2} ((16 - x^4) - (x^4 - 8x^2 + 16)) \, dx = \int_{-2}^{2} (-2x^4 + 8x^2) \, dx\]

Step 5: Evaluate the Integral

Calculate the integral: \[\int_{-2}^2 (-2x^4 + 8x^2) \, dx = \left[ \frac{-2x^5}{5} + \frac{8x^3}{3} \right]_{-2}^{2}\]Evaluate this from \(-2\) to \(2\).

Step 6: Compute the Final Result

Plug in the limits to the evaluated integral: \[\left( \frac{-2(2)^5}{5} + \frac{8(2)^3}{3} \right) - \left( \frac{-2(-2)^5}{5} + \frac{8(-2)^3}{3} \right)\]Simplify and calculate the result.

Key Concepts

Integration

Integration is a key mathematical operation that allows us to calculate the area under curves, among other things. In this context, when examining the area between two curves, integration helps to find the enclosed area. We need to understand both the function that defines the curves and the limits between which the integration occurs. For the problem at hand, you perform integration to compute the definite integral, which provides the exact area between the given curves. The integration process deals with summing small areas underneath the curve over a defined interval. Here, integration is performed between the points where the two functions intersect, which form the limits of integration.

• Integration provides a sum of infinitesimally small quantities.
• It computes areas under or between curves over a specific interval.
• A definite integral, specifically, gives a numerical area measurement bounded by these limits.

Points of Intersection

Determining the points of intersection between two curves is an important step in solving problems involving areas bounded by those curves. These points indicate where the graphs of the functions meet and thus mark the limits of integration. Finding these points involves setting the two function equations equal to each other, resulting in a new equation that can be solved for the variable typically denoted as "x." For the exercise in question, when you set the two functions equal, it simplifies to a polynomial equation. Solving this equation yields the x-values where the curves intersect.

• Solving for points of intersection helps establish the boundaries for integration.
• These points tell you where one graph transitions to being "on top" of the other.
• The intersection points form the integration limits, vital for computing the bounded area correctly.

Definite Integral

A definite integral represents the accumulation of quantities, such as area, between two boundaries. In the context of areas bounded by curves, a definite integral calculates the total area between two intersecting curves over a specified interval. Consider the problem that involves integrating the function obtained from subtracting one curve's equation from the other. After determining the limits of integration from the intersection points, we compute the definite integral. This integral essentially accumulates the differences between the two curves over the given range, providing the enclosed area.

It is evaluated over a specific interval.

Provides exact numerical values for area calculations.

In the formula \( \int_a^b (y_1 - y_2) \, dx \), \( y_1 \) and \( y_2 \) are the bounding curves.

This process involves taking antiderivatives, substituting limits, and finally subtracting, to provide the required area value.