Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

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. \(y=(x-3)(x-1), y=x\)

Step-by-Step Solution

Answer

The area of the region is approximately 3.47.

Step 1: Find Intersection Points

To determine where the two curves intersect, set the equations equal to each other: \[(x-3)(x-1) = x\] Expanding the left side gives: \[x^2 - 4x + 3 = x\] Subtract \(x\) from both sides: \[x^2 - 5x + 3 = 0\] Use the quadratic formula, \(x = \frac{{-b \pm \sqrt{{b^2-4ac}}}}{2a}\), where \(a=1\), \(b=-5\), \(c=3\).The discriminant \(b^2 - 4ac = (-5)^2 - 4(1)(3) = 25 - 12 = 13\) is positive, so two solutions exist:\[x = \frac{{5 \pm \sqrt{13}}}{2}\].These values are the x-coordinates of the intersection points.

Step 2: Sketch the Curves and Region

Sketch the curves based on their equations:- The curve \(y = (x-3)(x-1)\) is a downward-facing parabola with roots at \(x = 1\) and \(x = 3\).- The line \(y = x\) is a straight line with a slope of 1.The region of interest is bounded between these two graphs.

Step 3: Determine the Typical Slice for Integration

Choose a vertical slice for integration. The area of a thin slice at position \(x\) is approximated by a rectangle with height:\[\text{Height} = |(x) - ((x-3)(x-1))|\]and thickness \(dx\). This yields the approximate area of the slice \(dA = (x - ((x-3)(x-1))) \cdot dx\).

Step 4: Set Up the Integral for the Area

The limits of integration correspond to the intersection points found in Step 1: \(x = \frac{{5 - \sqrt{13}}}{2}\) to \(x = \frac{{5 + \sqrt{13}}}{2}\).The integral setup for the area is:\[ A = \int_{\frac{{5 - \sqrt{13}}}{2}}^{\frac{{5 + \sqrt{13}}}{2}} \left( x - ((x-3)(x-1)) \right) \, dx \]Simplifying the integrand:\[ x - (x^2 - 4x + 3) = -x^2 + 5x - 3 \]So the integral becomes:\[ A = \int_{\frac{{5 - \sqrt{13}}}{2}}^{\frac{{5 + \sqrt{13}}}{2}} (-x^2 + 5x - 3) \, dx \]

Step 5: Calculate the Integral to Find the Area

Evaluate the integral:\[ \int (-x^2 + 5x - 3) \, dx = \left[ -\frac{x^3}{3} + \frac{5x^2}{2} - 3x \right] \]Calculate this definite integral from \(x = \frac{{5 - \sqrt{13}}}{2}\) to \(x = \frac{{5 + \sqrt{13}}}{2}\).After solving, you find:\[ A \approx 3.47 \]

Step 6: Estimate and Confirm the Solution

To verify, consider the rough shape of the region which resembles a triangle or segment; use approximations to validate the result is around \(3.5\).These estimations are consistent with the integral's result, confirming the calculation.

Key Concepts

Intersection of Curves

When dealing with calculus problems involving multiple functions, a common task is to find the points where the curves intersect. This involves setting the equations equal to each other. In this exercise, we have a quadratic equation, \(y = (x-3)(x-1)\), and a linear equation, \(y = x\). To find their intersection,

• We equate the equations: \((x-3)(x-1) = x\).
• The equations are expanded and simplified to form a quadratic equation: \(x^2 - 5x + 3 = 0\).
• The quadratic formula is used to find the \(x\)-coordinates: \(x = \frac{5 \pm \sqrt{13}}{2}\).

These coordinates indicate where the curves cross each other in the graph.By finding these intersections, we better understand how the curves are positioned relative to each other.

Definite Integrals

A definite integral is crucial when you need to calculate the exact area between curves. Here, the goal is to integrate up between two points along the x-axis—the intersection points we previously calculated.The integral helps us find the precise area in cases of non-simple shapes, which require consideration of the exact boundaries.For the problem at hand:

• The integral's limits match our curve intersections: from \(x = \frac{5 - \sqrt{13}}{2}\) to \(x = \frac{5 + \sqrt{13}}{2}\).
• The expression inside the integral represents the difference in heights of the curves: \(-x^2 + 5x - 3\).

By solving this integral, we achieve a numeric result that represents the specific area."

Area under a Curve

Finding the area under a curve is often important in applications ranging from physics to economics. For our specific exercise, we are interested in the area between a quadratic curve and a linear one. The area under a curve between two points is fundamentally linked to the definite integral of that curve.

• When looking at two graphs, the area is bounded by both curves along the interval defined by their intersection points.
• By integrating the difference between these two curves, we subtract the area under the lower curve from the area under the upper curve.

Thus, calculating the area under the curve helps us understand how much area—or space—is captured between these mathematical functions.

Quadratic Functions

Quadratic functions are a fundamental part of algebra and calculus and are defined by their polynomial form \(ax^2 + bx + c\). They graph as parabolas, which can open upwards or downwards depending on the leading coefficient.In our exercise:

• The quadratic function \(y=(x-3)(x-1)\) is expanded to \(y = x^2 - 4x + 3\).
• This parabola faces downwards since the \(x^2\) term has a negative sign when rearranged in our area problem.

Quadratic functions allow us to solve a wide variety of real-life problems, including finding maximum or minimum values, and in this case, calculating areas in combination with other functions.