Problem 40 - Calculus Early Transcendentals

Question

Sketch the region and find its area. The region bounded by \(y=\sin x\) and \(y=x(x-\pi),\) for \(0 \leq x \leq \pi\)

Step-by-Step Solution

Answer

Question: Find the area of the region bounded by the two functions \(y=\sin x\) and \(y=x(x-\pi)\) for the domain \(0 \leq x \leq \pi\). Answer: The area of the region bounded by the two functions is approximately 2.2019 square units.

Step 1: Analyzing the Functions

First, let's understand the two functions. 1. \(y=\sin x\) is the sine function, which is a periodic function oscillating between -1 and 1. 2. \(y=x(x-\pi)\) is a quadratic function, opening upwards with roots at \(x=0\) and \(x=\pi\).

Step 2: Finding Intersection Points and Sketching the Graphs

To find the intersection points of the two functions, we need to solve the equation: \(\sin x = x(x-\pi)\). We can see that \(x=0\) is a solution: \(y = \sin(0) = 0\) \(y = 0(0-\pi) = 0\) Now for other intersection points, we need to solve the equation \(\sin x = x(x-\pi)\) for \(0 < x < \pi\). Unfortunately, there is no algebraic method to solve this equation. We need to rely on graphical methods or numerical methods (such as the bisection method or Newton's method) to approximate the solution. Let us call the intersection point as \(x = x_1\). Now considering the domain \(0 \leq x \leq \pi\), we can sketch the graph of both functions to visualize the region we need to find the area of.

Step 3: Sketching the Region

Here is the sketch of the two functions and the region bounded by the two functions, with the two intersection points labeled as \(A (0,0)\) and \(B(x_1, y_1)\). [Graph not provided]

Step 4: Finding the Area of the Region

Now that we have the intersection points and the sketch of the bounded region, we can find the area by performing the integral of the difference between the two functions. The area of the bounded region can be represented as: \(Area = \int_0^{x_1} (x(x-\pi) - \sin x) \, dx\)

Step 5: Evaluate the Integral and Find the Area

Let's evaluate the integral to find the area. First, we will integrate each of the functions separately: 1. \(\int_0^{x_1} x(x-\pi) \, dx = \int_0^{x_1} (x^2 - \pi x) \, dx\) Integrating term by term, we obtain: \(\frac{1}{3}x^3 - \frac{\pi}{2}x^2 \Big|_0^{x_1}\) 2. \(\int_0^{x_1} \sin x \, dx\) The antiderivative of \(\sin x\) is \(-\cos x\). Evaluating, we get: \(-\cos x \Big|_0^{x_1}\) Now, calculating the difference: \(Area = \left(\frac{1}{3}x_1^3 - \frac{\pi}{2}x_1^2\right) - (-\cos x_1 + 1)\) From the sketch, we can see that there is only one intersection point and also we can conclude that \(0 < x_1 < \pi\). Using any numerical method like Newton's method, we can approximate \(x_1 \approx 2.0288\). Plug in this value into the equation and evaluate it numerically to find the area. Therefore, the area of the region bounded by the two functions is approximately 2.2019 square units.

Key Concepts

Integration

Integration is a process that allows us to find the area under a curve or between curves in a given interval. When working with functions like in our exercise, we focus on integrating the difference between the functions' values over specified limits.
Integration rounds up individual values over an interval to provide us with the accumulated value, which represents the area in this context.
There are a few key steps to integrating:

• Identify the limits of the area you are interested in; these are often your intersection points.
• Determine your functions, which you'll be working with.
• Set up your integral by subtracting the lower function from the upper function within the given limits.
• Perform the integration to determine the result, which gives the area of the region bounded by the curves.

This method works beautifully when you have algebraic expressions, as long as the functions involved are integrable over the interval.

Intersection Points

Finding intersection points is a crucial step because they define where two curves meet. This information is essential when determining the bounded area.
In our scenario with functions like the sine and a quadratic, these points show us where one function overtakes the other within a given section of the graph.
The general process is as follows:

• Equate the values of the two functions: for example, \sin x = x(x - \pi) for our exercise.
• Solve for \(x\), though this may not always be possible algebraically. Sometimes the solution requires numerical approximation.
• Recognize these points graphically or through numerical methods; they represent the limits of integration.

Understanding where two functions intersect aids in correctly setting up the integral limits.

Definite Integrals

Definite integrals give us the precise area between a curve and a section of the x-axis within a specific interval. When working with definite integrals, you calculate the difference in the value of an integral function at specified limits.
For example, using our exercise:
You calculate:

• Integrate each function separately over the interval defined by the intersection points.
• Subtract the integrated values of the sine function from the quadratic function.

This process reveals the exact space the curves enclose. When dealing with areas, the construction of the definite integral considers the signs of the areas processed; it handles overlapping and negates any negative values resulting from the subtraction, focusing only on the absolute values that represent physical area.

Numerical Methods

When algebraic solutions aren't suitable for finding precise solutions, numerical methods step in. These methods approximate solutions such as finding the intersection points or evaluating an integral.
Numerical techniques come in handy for exercises like this one, where exact solutions are elusive.
Some common numerical methods include:

• Bisection method: A technique for finding roots by dividing intervals and checking where the sign changes.
• Newton's method: Utilizes tangent lines for successive approximations, quickly zeroing in on precise numerical answers.

By using these methods, integration becomes possible even when manual calculations would be infeasible. In our task, such methods help to approximate the intersection point \(x_1\) accurately, ensuring we precisely calculate the bounded area between the curved outlines.