Problem 25 - Single Variable Calculus

Question

In Exercises \(9-40\), sketch the region bounded by the graphs of the given equations and find the area of that region. $$ y=-x^{2}+6 x+5, \quad y=x^{2}+5 $$

Step-by-Step Solution

Answer

The points of intersection between the functions \(y=-x^2+6x+5\) and \(y=x^2+5\) are at (0, 5) and (3, 14). To find the area of the region bounded by these functions on the interval [0, 3], we integrate the difference of the two functions, resulting in \(A = \int_0^3 (2x^2 - 6x) \,dx\). After integrating and calculating the result, we obtain an area of 9 square units.

Step 1: Find the points of intersection between the two functions.

To find the points of intersection, we need to find the values of \(x\) for which both functions are equal. We'll set them equal to each other and solve for \(x\). $$ -x^2 + 6x + 5 = x^2 + 5 $$ Combine the like terms to get a quadratic equation: $$ 2x^2 - 6x = 0 $$ Now, factor the equation to find the values of \(x\): $$ 2x(x - 3) = 0 $$ So, the points of intersection are at \(x = 0\) and \(x = 3\). To find their corresponding y-values, we can plug these x-values into either of the original functions. We will use \(y = x^2 + 5\) for simplicity: For x = 0: $$ y = 0^2 + 5 = 5 $$ For x = 3: $$ y = 3^2 + 5 = 14 $$ Thus, the points of intersection are (0, 5) and (3, 14).

Step 2: Calculate the area of the region between the two functions using definite integration.

To find the area of the region bounded by the graphs of the given equations, we need to integrate the difference of the two functions on the interval [0, 3]. We'll take the "upward" parabola and subtract the "downward" parabola: $$ A = \int_0^3 (x^2 + 5 - (-x^2 + 6x + 5)) \,dx $$ Combine the like terms: $$ A = \int_0^3 (2x^2 - 6x) \,dx $$ Now integrate: $$ A = \left[ \frac{2}{3}x^3 - 3x^2 \right]_{0}^{3} $$ Substitute the limits of integration: $$ A = \left( \frac{2}{3}(3)^3 - 3(3)^2 \right) - \left( \frac{2}{3}(0)^3 - 3(0)^2 \right) $$ Calculate the result: $$ A = \left( \frac{54}{3} - 27 \right) = 18 - 27 = -9 $$ The result is negative but we are looking for the area, so we take the absolute value: $$ A = |-9| = 9 $$

Step 3: Sketch the region bounded by the graphs of the given equations.

To sketch the region bounded by the graphs, we can draw the two parabolas and shade in the area between them from x = 0 to x = 3: 1. Sketch the upward parabola: \(y = x^2 + 5\) 2. Sketch the downward parabola: \(y = -x^2 + 6x + 5\) 3. Shade the region between the two parabolas from x = 0 to x = 3. 4. Label the points of intersection, (0, 5) and (3, 14). The area of the bounded region is 9 square units.

Key Concepts

Definite Integration

Definite integration helps us find the area between two curves over a specific interval. In this exercise, we explore how to calculate the area between two parabolic curves from their intersection points. The process involves integrating the difference between the two functions. Here’s a quick breakdown of the steps:

• Identify the interval by locating the points of intersection.
• Set up the integral by subtracting the equations of the curves. The upper function comes first.
• Integrate over the defined interval to find the net area.
• Take the absolute value if the result of the definite integral is negative, as we are calculating area, which is always positive.

By integrating the difference from 0 to 3, we found the area between the parabolas.

Parabolas

Parabolas are symmetrical, U-shaped graphs that can open upwards or downwards. They are represented by quadratic equations and are key in this exercise.

The given functions represent an upward-opening and a downward-opening parabola:

• The upward parabola: \(y = x^2 + 5\)
• The downward parabola: \(y = -x^2 + 6x + 5\)

To visualize how these parabolas create the bounded area, consider their orientations:

- The upward parabola has its vertex at the minimum point and widens as x increases.
- The downward parabola, conversely, has its vertex at the maximum point.

The area between them is what we calculate using definite integration.

Points of Intersection

Points of intersection tell us where the two curves meet. These points form the boundaries of the area we need to find.

Finding these points involves:

• Setting the equations equal: \(-x^2 + 6x + 5 = x^2 + 5\)
• Solving the resulting quadratic equation: \(2x^2 - 6x = 0\)
• Factoring to find the solutions for \(x\): \(x = 0\) and \(x = 3\)

These values signify the x-coordinates where the two functions intersect. Plug these back into either of the original functions to find the y-coordinates, resulting in the points (0, 5) and (3, 14). These points are crucial as they define the interval for integration, providing the limits from x = 0 to x = 3.

Quadratic Equations

Quadratic equations form the foundation of parabolas and are vital for graphing and intersection analysis.

Each quadratic equation can be written in the standard form: \(ax^2 + bx + c = 0\). In this exercise:

• The downward parabola is given by \(y = -x^2 + 6x + 5\)
• The upward parabola by \(y = x^2 + 5\)

Solving for points of intersection involves operations on these equations. By rearranging and combining terms:

- We formulated: \(2x^2 - 6x = 0\)
- Factored it as: \(2x(x - 3) = 0\)

This shows the simplicity and symmetry of quadratics, allowing us to solve efficiently by finding x-values where two parabolas meet.