Problem 41 - Calculus Early Transcendentals

Question

Sketch the region and find its area. The region bounded by \(y=(x-1)^{2}\) and \(y=7 x-19\)

Step-by-Step Solution

Answer

Answer: The area of the region bounded by the functions is approximately 100.3 square units.

Step 1: Sketch the graphs

Start by sketching the graphs of the two functions \(y=(x-1)^{2}\) and \(y=7x-19\). The first function is a parabola with its vertex at \((1,0)\) and opens upward, while the second function is a straight line with a slope of 7 and a y-intercept of -19. It might be useful to create a table of values for both functions to help with the sketching.

Step 2: Find the intersection points

To find the intersection points, set the two functions equal to each other and solve for \(x\) \((x-1)^{2} = 7x - 19\). Expanding and moving the terms to one side will create a quadratic equation, which we'll solve to find the x-values of the intersection points. \(x^{2} -2x + 1 = 7x - 19\) \(x^{2} - 9x + 20 = 0\) Factor the quadratic or use the quadratic formula to find the x-values. In this case, factoring gives: \((x-4)(x-5) = 0\) The x-values for the intersection points are \(x=4\) and \(x=5\). Plug these x-values into either function to find their respective y-values. \(y_1=(4-1)^2 = 9\), \(y_2=(5-1)^2 = 16\). The intersection points are \((4,9)\) and \((5,16)\).

Step 3: Set up the integral for the area

To find the area between the two curves, we want to subtract the function values, then integrate over the interval between the intersection points. The area between the curves can be found using the following integral: \(\int_{4}^{5} (7x-19) - (x-1)^{2} dx\)

Step 4: Evaluate the integral

Now, evaluate the integral: \(\int_{4}^{5} (7x-19) - (x^2-2x+1) dx\) Combine like terms within the integral: \(\int_{4}^{5} -x^2 + 9x - 20 dx\) Find the antiderivative: \(-\frac{1}{3}x^3 + \frac{9}{2}x^2 - 20x \bigg|_4^5\) Evaluate the antiderivative at the bounds of integration and subtract: \(\left(-\frac{1}{3}(5)^3 + \frac{9}{2}(5)^2 - 20(5) \right) - \left(-\frac{1}{3}(4)^3 + \frac{9}{2}(4)^2 - 20(4) \right)\) Calculate the final result: \begin{align*} \left(-\frac{1}{3}(125)+\frac{9}{2}(25) - 100 \right) - \left(-\frac{1}{3}(64) + \frac{9}{2}(16) - 80 \right) &= \left(-41.\overline{6} + 112.5 - 100\right) - \left(-21.\overline{3} + 72 - 80\right) \\ &= 71 - (-29.3\overline{3}) \\ &= 71 + 29.3\overline{3} \\ &\approx 100.\overline{3} \end{align*}

Step 5: State the final answer

The area of the region bounded by the functions \(y=(x-1)^{2}\) and \(y=7x-19\) is approximately 100.3 square units.

Key Concepts

Understanding Integrals

Integrals are a significant concept in calculus, providing solutions to problems involving areas, accumulation, and more. Think of an integral as a tool for adding things up. In geometry, this often means finding the area under a curve or between curves. This concept is crucial for understanding how to handle situations where you need the cumulative total of a varying quantity.

When you're looking at a function on a graph, the integral helps you calculate the total area underneath the function over a particular interval. This is known as the definite integral. The general form of the integral is represented as:

• \int_{a}^{b} f(x)\,dx

This expression denotes the integral of the function \( f(x) \) from \( a \) to \( b \). In simple terms, it means to add up the area underneath \( f(x) \) starting at \( x = a \) and ending at \( x = b \). Understanding this concept will greatly aid in calculating areas, such as the area between different curves.

Finding Area Between Curves

To find the area between curves, we use integrals, specifically the concept of finding the area under one curve and subtracting it from the area under another. This is quite common when dealing with two intersecting graphs, like a line and a parabola. In our example, the functions \( y = (x-1)^2 \) and \( y = 7x-19 \) create such a scenario.

Here's how to find the area between them:

First, determine where the curves intersect, as this will be your limits of integration \( a \) and \( b \). Calculate the integral of both functions separately, then subtract the smaller area from the larger area. This gives the total space between the functions. The relevant formula here is:

• \int_{a}^{b} [f(x) - g(x)]\,dx

where \( f(x) \) is the upper function and \( g(x) \) is the lower function. This approach ensures you effectively find the difference in areas, capturing the region you are interested in.

Quadratic Equations and Their Role

Quadratic equations are polynomials of degree 2, generally taking the form \( ax^2 + bx + c = 0 \). These equations graph as parabolas and can open upwards or downwards depending on the coefficient \( a \). Recognizing these properties helps determine the shape and position of the curve when sketching.

For the exercise at hand, the function \( y = (x-1)^2 \) is a quadratic equation. Solving quadratic equations often involves finding their roots, which provides important information about the graph, such as vertex and intersections with the x-axis.

We often solve them using:

• Factoring
• Completing the square
• Quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Choosing the right method depends on the specific problem. In our scenario, factoring was sufficient to find the intersection points, leading us to understand how the curves intersected.

Calculating Intersection Points

Intersection points are where two graphs meet, which is crucial for determining the area between curves. Finding these involves setting the functions equal and solving for \( x \). These points not only serve as limits for integration but also help in sketching the region more accurately.

In our exercise, we set \( (x-1)^2 = 7x - 19 \) to determine where the parabola and the line intersect. After expanding and rearranging to form the quadratic equation \( x^2 - 9x + 20 = 0 \), we solve for \( x \) by factoring, giving \( x = 4 \) and \( x = 5 \).

By substituting back into either function, we also confirm the corresponding \( y \) values for these \( x \) values, which yield the precise coordinates of the intersection points: \( (4, 9) \) and \( (5, 16) \). This calculation is crucial, as accurately identifying these points determines the correct boundaries for evaluating the integral.