Problem 19 - 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: \(y=(x-3)(x-1), y=x\)

Step-by-Step Solution

Answer

The area of the region is \(\approx 1.385\) square units.

Step 1: Sketch the Graphs

First, we sketch the graphs of the two given equations: \(y = (x-3)(x-1)\) and \(y = x\). The graph of \(y = (x-3)(x-1)\) is a downward opening parabola, intersecting the x-axis at \(x = 1\) and \(x = 3\). The graph of \(y = x\) is a straight line crossing through the origin and has a slope of 1. Find their points of intersection by equating them: \((x-3)(x-1) = x\).

Step 2: Find Points of Intersection

To find the points of intersection, solve the equation \((x-3)(x-1) = x\). Expand and simplify this quadratic equation: \(x^2 - 4x + 3 = x\). Rearrange it to \(x^2 - 5x + 3 = 0\). Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a=1, b=-5, c=3\). Solving gives two solutions: \(x = \frac{5 \pm \sqrt{13}}{2}\).

Step 3: Identify the Bounded Region

The bounded region is between \(y = (x-3)(x-1)\) and \(y = x\) from \(x = \frac{5 - \sqrt{13}}{2}\) to \(x = \frac{5 + \sqrt{13}}{2}\). The parabola is below the line within this interval.

Step 4: Set Up the Integral for Area

Set up the integral to find the area between the curves \(y = x\) and \(y = (x-3)(x-1)\). The integral is: \ \(\int_{\frac{5-\sqrt{13}}{2}}^{\frac{5+\sqrt{13}}{2}} \left(x - [(x-3)(x-1)]\right) \, dx\).

Step 5: Simplify the Integrand

Simplify the integrand: \(x - (x^2 - 4x + 3)\) simplifies to \(-x^2 + 5x - 3\). This is the function to integrate.

Step 6: Integrate

Integrate \(-x^2 + 5x - 3\) from \(\frac{5-\sqrt{13}}{2}\) to \(\frac{5+\sqrt{13}}{2}\). This gives: \ \(\left[ -\frac{x^3}{3} + \frac{5x^2}{2} - 3x \right]_{\frac{5-\sqrt{13}}{2}}^{\frac{5+\sqrt{13}}{2}} \).

Step 7: Evaluate the Definite Integral

Substitute in the limits: compute \(-\frac{\left(\frac{5+\sqrt{13}}{2}\right)^3}{3} + \frac{5\left(\frac{5+\sqrt{13}}{2}\right)^2}{2} - 3\left(\frac{5+\sqrt{13}}{2}\right)\) and do the same for \(\frac{5-\sqrt{13}}{2}\). Subtract these results to find the area.

Step 8: Check with an Estimate

Estimate the bounded area using geometric approximations or numerical methods. By comparing with the calculated integral, ensure the results are close. This confirms accuracy.

Key Concepts

Integral Calculus

Integral calculus is an essential branch of calculus that focuses on the concept of integration.It plays a critical role in calculating areas, volumes, and other quantities under curves.Here, we use integrals to find the area between two curves.When we talk about integrals, we're referring to the process of finding the function that describes the accumulation of quantities.In this context, the definite integral is key, as it calculates the total area under a curve within specified bounds.The basic notation for an integral is \[\int_{a}^{b} f(x) \, dx\]where:

• \(f(x)\) is the function being integrated,
• \(a\) and \(b\) define the interval of integration,
• \(dx\) indicates the variable of integration.

To solve the problem of finding the area between two given curves, you'll subtract one integral from another.This results in the net area between the curves.

Area under Curve

Finding the area under a curve is one of the primary applications of integral calculus.In this problem, we calculate the area bounded by two curves.The task is to determine how much space lies between the curves from one point of intersection to another.The area under the curve is essentially the integral of the function over the given interval.For area between two curves \(y_1\) and \(y_2\) from \(x = a\) to \(x = b\), the formula is:\[\int_{a}^{b} (y_1 - y_2) \, dx\]Here's the step-by-step approach to find this area:

• Identify the top and bottom functions over the interval.
• Integrate the difference of these functions over the specified interval.
• The result gives the net area between the curves.

This method helps you understand not just the language of mathematics, but how it can be visualized and measured in real-world scenarios.

Graph Sketching

Graph sketching is an invaluable skill in both calculus and broader mathematical contexts.It helps visualize functions and their interactions with each other, like finding intersections or seeing overlaps.For the current exercise, sketching the given functions, \(y = (x-3)(x-1)\) and \(y = x\), is a crucial first step.

• **Parabolic Graph**: The equation \(y = (x-3)(x-1)\) forms a downward-opening parabola.The roots of this equation at \(x = 1\) and \(x = 3\) help identify critical points, indicating where the parabola touches the x-axis.
• **Linear Graph**: The equation \(y = x\) is a straight line that passes through the origin with a slope of 1.This makes it simpler to visualize and sketch, as straight lines only require two points.

By sketching these graphs, you quickly visualize the "bounded region" needed for further calculation.

Intersection Points

Intersection points of graphs occur where functions have the same values at given points.Determining these is essential for defining the interval over which you'll calculate the integral to find the area between curves.To find intersections for the curves given by \(y = (x-3)(x-1)\) and \(y = x\), you equate them:\[(x-3)(x-1) = x\]Upon simplifying, you get a quadratic equation:\[x^2 - 5x + 3 = 0\]This can be solved using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]With coefficients \(a=1\), \(b=-5\), \(c=3\), the solution identifies the intersection points:\[x = \frac{5 \pm \sqrt{13}}{2}\]These points are crucial as they define the lower and upper limits of the definite integral used to find the area between the curves.