Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 11-30, sketch the region bounded by the graphs of the given equations, show a typical slice, approximate its area, set up an integral, and calculate the area of the region. Make an estimate of the area to confirm your answer: \(y=x^{2}-4 x-5, y=0\), between \(x=-1\) and \(x=4\)

Step-by-Step Solution

Answer

The area is \( \frac{100}{3} \) square units.

Step 1: Understand the Functions

The equations provided are \[ y = x^2 - 4x - 5 \] and \( y = 0 \). First, let's interpret these functions. The equation \( y = x^2 - 4x - 5 \) describes a quadratic function, which will be a parabola opening upwards, due to the positive coefficient of \( x^2 \). The equation \( y = 0 \) represents the x-axis.

Step 2: Find Intersection Points

To understand the boundaries, we need to find the intersection points of the quadratic function with the x-axis. Set the quadratic equal to zero: \[ x^2 - 4x - 5 = 0 \]Solve this equation using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1 \), \( b = -4 \), and \( c = -5 \). This gives roots:\[ x = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm \sqrt{36}}{2} = \frac{4 \pm 6}{2} \]Therefore, the roots are \( x = 5 \) and \( x = -1 \). These are the x-intercepts.

Step 3: Define the Region of Interest

The region of interest is between the x-values where the parabola intersects the x-axis, limited to \( x = -1 \) and \( x = 4 \). We are interested in the area from \( x = -1 \) to \( x = 4 \).

Step 4: Approximate the Area

To visualize the area, draw a typical vertical slice through the region between \( x = -1 \) and \( x = 4 \). This slice represents a thin rectangle, where the height is given by the function value \( y = x^2 - 4x - 5 \) at given x, and the width is a small change in \( x \), often denoted \( \Delta x \). Estimate the area by summing these rectangles.

Step 5: Set Up the Integral

The exact area is found using an integral. Set up the definite integral of the function from the lower to upper bounds of \( x = -1 \) to \( x = 4 \):\[ \int_{-1}^{4} (x^2 - 4x - 5) \, dx \]

Step 6: Calculate the Integral

Calculate the integral:\[ \int (x^2 - 4x - 5) \, dx = \frac{x^3}{3} - 2x^2 - 5x + C \]Evaluate this from \( x = -1 \) to \( x = 4 \):\[ \left[ \frac{x^3}{3} - 2x^2 - 5x \right]_{-1}^{4} = \left( \frac{4^3}{3} - 2(4)^2 - 5(4) \right) - \left( \frac{(-1)^3}{3} - 2(-1)^2 - 5(-1) \right) \]\[ = \left( \frac{64}{3} - 32 - 20 \right) - \left( \frac{-1}{3} - 2 + 5 \right) \]\[ = \left( \frac{64}{3} - \frac{156}{3} \right) - \left( \frac{-1}{3} + \frac{9}{3} \right) \]\[ = \frac{-92}{3} + \frac{8}{3} \]\[ = \frac{-100}{3} = -\frac{100}{3} \]The negative sign indicates the area is below the x-axis. So, the area is \( \frac{100}{3} \).

Step 7: Confirming with Estimation

Estimate by considering the shape's symmetry and rough geometry over the interval. Observing from the graph, the symmetric nature of the parabola suggests an area close to our calculation. This confirms the precise integration approach matched visual estimation.

Key Concepts

Definite Integrals

A definite integral is a way of finding the area under a curve between two points. In this exercise, we are finding the area under the curve of a quadratic function \( y = x^2 - 4x - 5 \), between \( x = -1 \) and \( x = 4 \). This involves setting up an integral from one x-value to another and solving it to find the exact area. Definite integrals consider not just the shape of the curve, but also the portion of the x-axis it spans.

• The limits of integration (\( x = -1 \) and \( x = 4 \)) define the interval over which the area is calculated.
• The integral calculates the "net" area, meaning it accounts for areas below the x-axis as well.

This is why solving the integral provides \( \frac{100}{3} \), which is the unsigned area between the parabola and the x-axis within the specified limits.

Quadratic Functions

Quadratic functions are algebraic expressions of the form \( y = ax^2 + bx + c \). They graph as parabolas, which can open upwards or downwards depending on the sign of \( a \). In this problem, the quadratic function \( y = x^2 - 4x - 5 \) is considered.

• The function has a positive \( a \) coefficient, indicating the parabola opens upwards.
• The standard form helps identify the vertex, axis of symmetry, and direction of the parabola.

To find points of interest like roots or x-intercepts, set the quadratic equal to zero and solve the equation. Using the quadratic formula helps find these intercepts efficiently. In this exercise, the roots \( x = 5 \) and \( x = -1 \) indicate where the parabola crosses the x-axis.

Area under Curve

The area under a curve represents the integral of the function over the given interval. It effectively measures the 'space' between the curve and the x-axis.

• This exercise involves a quadratic curve, and the area under it is calculated by integrating the function between specified x-values.
• An important consideration is whether parts of the curve are above or below the x-axis, which affects the integral's sign.

For \( y = x^2 - 4x - 5 \), the value of the definite integral provides a measure of \(-\frac{100}{3}\), which when taking absolute value gives the total area of the region bounded by the curve, ensuring meaningful results even though part of the area is below the x-axis. The deduction involves finding the mathematical "net" area and interpreting its meaning in terms of the curve's geometry.

Graphing Functions

Graphing functions helps visualize mathematical concepts clearly. In this context, graphing the quadratic function \( y = x^2 - 4x - 5 \) aids in understanding its behavior, including its intercepts and general shape.

• The roots or x-intercepts \( x = 5 \) and \( x = -1 \) can be clearly plotted to visualize where the function crosses the x-axis.
• The graph's overall structure shows a parabola opening upwards, which is important for visualizing the region whose area is being calculated.

Plotting such functions can also facilitate verifying analytical solutions—when the parabola is sketched, it becomes clear whether calculated areas or intercepts are reasonable, improving comprehension and problem-solving accuracy.

Roots of Equations

Finding the roots of equations is crucial for solving and understanding quadratic functions. In this exercise, determining the roots of \( y = x^2 - 4x - 5 \) (by setting it equal to zero) finds the points where the function intersects the x-axis.

• The quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), is applied to find these roots.
• For this function, the calculations yield \( x = 5 \) and \( x = -1 \).

Roots are critical as they define the interval of integration. Understanding where these intersections occur can guide setup in an integral problem, dictating bounds of integration critical to compute areas accurately, as demonstrated by the definite integral in this exercise.