Problem 22 - Calculus Single Variable

Question

Find the area of the region that is bounded by the graphs of \(y=f(x)\) and \(y=g(x)\) for \(x\) between the abscissas of the two points of intersection. $$ f(x)=2 x+3 \quad g(x)=9+x-x^{2} $$

Step-by-Step Solution

Answer

The area between the curves is the definite integral \( \int_{-3}^{2} (9 + x - x^2) - (2x + 3) \, dx \).

Step 1: Find Points of Intersection

To find the points of intersection of the graphs of \( y = f(x) \) and \( y = g(x) \), we set the equations equal:\[ 2x + 3 = 9 + x - x^2 \]Rearranging terms gives us:\[ x^2 + x - 6 = 0 \].We solve this quadratic equation using the quadratic formula \( x = \frac{{-b \, \pm \, \sqrt{{b^2 - 4ac}}}}{2a} \), where \( a = 1 \), \( b = 1 \), and \( c = -6 \).

Step 2: Solve the Quadratic Equation

Using the quadratic formula:\[ x = \frac{{-1 \, \pm \, \sqrt{{1^2 - 4 \times 1 \times (-6)}}}}{2 \times 1} \]\[ x = \frac{{-1 \, \pm \, \sqrt{{1 + 24}}}}{2} \]\[ x = \frac{{-1 \, \pm \, 5}}{2} \].This gives the solutions \( x_1 = 2 \) and \( x_2 = -3 \).

Step 3: Set Up the Integral for the Area

The area between the curves from \( x = -3 \) to \( x = 2 \) is given by the integral of the difference of the functions:\[ \int_{x=-3}^{x=2} (g(x) - f(x)) \, dx \]Substitute \( g(x) = 9 + x - x^2 \) and \( f(x) = 2x + 3 \) into the equation:\[ \int_{x=-3}^{x=2} ((9 + x - x^2) - (2x + 3)) \, dx \].

Key Concepts

Points of Intersection

The process of finding points of intersection is essential in calculating the area between two curves. It involves determining where the graphs of two equations meet. This is done by setting the equations equal to each other. For instance, with equations like \( y = 2x + 3 \) and \( y = 9 + x - x^2 \), equating them results in the equation \( 2x + 3 = 9 + x - x^2 \).

This rearranges to a simplified form \( x^2 + x - 6 = 0 \). These points of intersection provide the limits of integration required for calculating the area between the curves. By solving the equation \( x^2 + x - 6 = 0 \), we determine the two intersection points \( x_1 = 2 \) and \( x_2 = -3 \).

Identifying intersection points is crucial because they show the interval on the x-axis where the area calculation should focus.

Quadratic Equation

A quadratic equation is a polynomial equation of degree 2, typically in the form \( ax^2 + bx + c = 0 \). Solving a quadratic equation can be done using various methods, including factoring, completing the square, or utilizing the quadratic formula.

In our context, we were tasked to solve \( x^2 + x - 6 = 0 \) to find the intersection points of two functions. This equation was solved using the quadratic formula \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \).

For this equation, \( a = 1 \), \( b = 1 \), and \( c = -6 \). Plugging these values into the formula gives:

• \( x = \frac{{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-6)}}}{2 \cdot 1} \)
• \( x = \frac{{-1 \pm 5}}{2} \)

Thus, the solutions are \( x_1 = 2 \) and \( x_2 = -3 \). These solutions not only confirm the points of intersection but also help in determining the limits for our integral.

Definite Integral

The definite integral plays a significant role in finding the area between curves. Once the points of intersection are found, the area between two curves can be calculated using the definite integral of the difference of the functions.

In this problem, we use the formula: \[ \int_{x=-3}^{x=2} (g(x) - f(x)) \, dx \]. This integral calculates the accumulated area between the curves from \( x=-3 \) to \( x=2 \). By substituting the expressions of the functions into this formula, we have:

• \( g(x) = 9 + x - x^2 \)
• \( f(x) = 2x + 3 \)

Substituting these into the integral gives:

• \[ \int_{x=-3}^{x=2} ((9 + x - x^2) - (2x + 3)) \, dx \]

The definite integral effectively calculates the net area, considering which function lies above the other in the given interval. By evaluating this integral, we can determine the exact area encased between the two curves.