Problem 64 - University Calculus: Early Transcendentals

Question

Find the areas of the regions enclosed by the lines and curves in Exercises \(63-72\). $$y=2 x-x^{2} \quad \text { and } \quad y=-3$$

Step-by-Step Solution

Answer

The area is 10.667 square units.

Step 1: Find Intersection Points

To find the limits of integration for the region bounded by the curve and the line, we need to find the intersection points of the equations. Set the equations equal to each other: \[ 2x - x^2 = -3 \]Rearrange the equation to find the intersection points:\[ x^2 - 2x - 3 = 0 \]Solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1 \), \( b = -2 \), and \( c = -3 \). This gives us:\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 1 \times (-3)}}{2} \]\[ x = \frac{2 \pm \sqrt{16}}{2} \]\[ x = \frac{2 \pm 4}{2} \]Thus, \( x = 3 \) and \( x = -1 \).

Step 2: Set Up the Integral for Area

The area between the curve \( y = 2x - x^2 \) and the line \( y = -3 \) can be found by integrating the difference between these two functions over the interval defined by their intersection points \([-1, 3]\).The integral setup is:\[ \text{Area} = \int_{-1}^{3} [(2x - x^2) - (-3)] \, dx \]Simplify the expression inside the integral:\[ \text{Area} = \int_{-1}^{3} [2x - x^2 + 3] \, dx \]

Step 3: Calculate the Integral

Now calculate the definite integral:\[ \text{Area} = \int_{-1}^{3} (2x - x^2 + 3) \, dx \]This involves integrating each term separately:\[ \int 2x \, dx = x^2 + C \]\[ \int -x^2 \, dx = -\frac{x^3}{3} + C \]\[ \int 3 \, dx = 3x + C \]So, the integral becomes:\[ \left[ x^2 - \frac{x^3}{3} + 3x \right]_{-1}^{3} \]Evaluate this expression at the bounds \(x = 3\) and \(x = -1\):At \(x = 3\):\[ 3^2 - \frac{3^3}{3} + 3(3) = 9 - 9 + 9 = 9 \]At \(x = -1\):\[ (-1)^2 - \frac{(-1)^3}{3} + 3(-1) = 1 + \frac{1}{3} - 3 = -1.667 \]Subtract these results:\[ 9 - (-1.667) = 10.667 \].

Step 4: Finalize the Area

The definite integral was calculated in the previous step and evaluated to 10.667. This result represents the area of the region enclosed by the given curve and line. Ensure that the units of the area are consistent with the context of the problem, if any units are specified.

Key Concepts

Quadratic Equation Solutions

A quadratic equation is any equation that can be rewritten in the standard form \( ax^2 + bx + c = 0 \) where \( a \), \( b \), and \( c \) are constants with \( a eq 0 \). These equations are important in calculus for finding intersection points between a curve and a line, as they define parabolic shaped graphs. To solve quadratic equations, one common technique is using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula provides the solutions (or roots) of the quadratic equation by determining the x-values where the polynomial equals zero. These x-values indicate where graphs intersect along the x-axis.

To use the quadratic formula, identify \( a \), \( b \), and \( c \) from the equation and substitute them into the formula. The expression under the square root, \( b^2 - 4ac \), is called the discriminant.

• If the discriminant is positive, there are two distinct real roots.
• If it's zero, there is one repeated real root.
• If it's negative, the roots are complex numbers.

This method was used in the exercise to solve \( x^2 - 2x - 3 = 0 \) to find points of intersection, which are critical in defining the limits for integration when finding areas between a curve and a line.

Definite Integral

In calculus, a definite integral is used to calculate the area under a curve between two points along the x-axis. It is represented by \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration and \( f(x) \) is the function being integrated. The process of integrating involves finding the antiderivative or integral of the function and then evaluating it at the two bounds.

To compute a definite integral:

• Find the antiderivative \( F(x) \) of the function \( f(x) \).
• Evaluate \( F(x) \) at the upper and lower bounds, \( b \) and \( a \).
• Subtract these values: \( F(b) - F(a) \).

The result provides the net area between the curve and the x-axis from \( a \) to \( b \). In the exercise, you used a definite integral
\[ \text{Area} = \int_{-1}^{3} (2x - x^2 + 3) \, dx \]
to determine the area enclosed by the curve \( y = 2x - x^2 \) and the line \( y = -3 \). This integral was evaluated to find the precise area by integrating the calculated function over the interval of the intersection points \([-1, 3]\).

Area Between Curves

Finding the area between curves involves more than one function and requires integration. Given two curves \( y = f(x) \) and \( y = g(x) \), where \( f(x) > g(x) \) over a particular interval, the area bound by them is found by integrating the difference of these functions over that interval. This is expressed as:
\[ \text{Area} = \int_{a}^{b} [f(x) - g(x)] \, dx \]
To compute this area:

• Determine the interval \([a, b]\) where the curves intersect by solving \( f(x) = g(x) \) to find bounds.
• Set up the integral with the integrand as \( f(x) - g(x) \).
• Evaluate the definite integral to find the area.

In the exercise, we were tasked to find the area enclosed between the curve \( y = 2x - x^2 \) and line \( y = -3 \). The process involved:
- Setting \( 2x - x^2 = -3 \) to find intersection points and form limits for integration.
- Integrating the function representing the area \([2x - x^2 + 3]\) between \( x = -1 \) and \( x = 3 \).
This approach illustrates how the area between two functions is calculated, providing a method to determine spaces enclosed by graphs.