Problem 6 - Calculus: Early Transcendentals

Question

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to \( x \) and \( y \). Draw a typical approximating rectangle and label its height and width. Then find the area of the region. \( y = \sin x \) , \( y = x \) , \( x = \frac{\pi}{2} \) , \( x = \pi \)

Step-by-Step Solution

Answer

Integrate with respect to \( x \), and the area is \( 1 - \frac{3\pi^2}{8} \).

Step 1: Identify the region of intersection

To find the region enclosed by the curves, we first need to determine their points of intersection. We solve for the intersection of the functions \( y = \sin x \) and \( y = x \) within the given bounds. This means solving \( \sin x = x \) for \( x \) between \( \frac{\pi}{2} \) and \( \pi \). However, within these limits, \( x = \sin x \) does not have any additional intersections except at the boundaries.

Step 2: Determine the boundaries of integration

The boundaries are defined by four curves: \( y = \sin x \), \( y = x \), \( x = \frac{\pi}{2} \), and \( x = \pi \). The intersection points and the vertical lines \( x = \frac{\pi}{2} \) and \( x = \pi \) confine the region.

Step 3: Decide whether to integrate with respect to x or y

Based on the curves and boundaries, it is easier to integrate with respect to \( x \) since the limits are given as vertical lines and the expression for \( y \) is simpler. Thus, we'll integrate from \( x = \frac{\pi}{2} \) to \( x = \pi \).

Step 4: Set up the integral

The height of each approximating rectangle is given by the difference in the \( y \)-values of the top function and the bottom function in this region. Here, \( y = \sin x \) is above \( y = x \), so the height is \( \sin x - x \). The width of each rectangle is \( dx \). Thus, the integral to find the area is: \[ A = \int_{\frac{\pi}{2}}^{\pi} (\sin x - x) \; dx \]

Step 5: Calculate the area

To compute the area, evaluate the definite integral: \[ A = \int_{\frac{\pi}{2}}^{\pi} (\sin x - x) \; dx = \left[ -\cos x - \frac{x^2}{2} \right]_{\frac{\pi}{2}}^{\pi} \]First compute for \( x = \pi \):\(-\cos(\pi) - \frac{\pi^2}{2} = -(-1) - \frac{\pi^2}{2} = 1 - \frac{\pi^2}{2} \)Then compute for \( x = \frac{\pi}{2} \):\(-\cos(\frac{\pi}{2}) - \frac{(\frac{\pi}{2})^2}{2} = 0 - \frac{\pi^2}{8} = -\frac{\pi^2}{8}\)Subtract these results:\(\left(1 - \frac{\pi^2}{2}\right) - \left(-\frac{\pi^2}{8}\right) = 1 - \frac{\pi^2}{2} + \frac{\pi^2}{8}\)Combine the terms:\( A = 1 - \frac{4\pi^2}{8} + \frac{\pi^2}{8} = 1 - \frac{3\pi^2}{8} \)

Key Concepts

Sketching Regions

When dealing with calculus and integration, sketching the region of interest is always a beneficial first step. It helps in visualizing the problem and understanding the confines of the area you're working with. In this exercise, we are sketching the region enclosed by the curves:

• \( y = \sin x \)
• \( y = x \)
• The vertical lines \( x = \frac{\pi}{2} \) and \( x = \pi \)

To sketch this region: - Plot the functions and vertical lines on a graph.- Notice that within the interval \( x = \frac{\pi}{2} \) to \( x = \pi \), these curves will intersect the boundaries, creating a closed shape. - Determine which part of the curves are above or below each other. Here, \( y = \sin x \) is above \( y = x \) within this interval. This sketch not only aids in determining the limits of integration but also in identifying the correct order of functions for calculating the area.

Definite Integrals

The concept of a definite integral is key in finding the area between curves. A definite integral calculates the signed area under a curve between two specific bounds, offering a precise interpretation of accumulated quantities. In this exercise, after identifying the enclosed region by sketching, you need to set up a definite integral. The bounds of the integral are from \( x = \frac{\pi}{2} \) to \( x = \pi \). This is determined by the given vertical lines:- The formula used for the integral set-up is: \[ A = \int_{\frac{\pi}{2}}^{\pi} (\sin x - x) \, dx \]- Here, \( \sin x - x \) defines the height of the rectangle, which is the difference between the values of the functions \( y = \sin x \) and \( y = x \). - \( dx \) is the thickness of these rectangles.Thus, the definite integral captures the entire shaded area in a single mathematical expression, allowing you to compute the area precisely.

Area Between Curves

Finding the area between curves is a fundamental application of calculus integration. It involves calculating the difference between two curves that enclose a region. In simpler terms, it is like layering one curve on top of another and measuring the space between them.In this example, the area to be calculated lies between the curves \( y = \sin x \) and \( y = x \) within the interval \( x = \frac{\pi}{2} \) to \( x = \pi \). Here is how you find this area:

• Identify the top and bottom functions within the interval – \( \sin x \) is the upper or top function.
• Subtract the bottom function from the top function to find the height of the typical rectangle (\( \sin x - x \)).
• Integrate this difference over the given interval.

Finally, you arrive at the formula to compute the area:\[ A = \int_{\frac{\pi}{2}}^{\pi} (\sin x - x) \, dx \]Once you evaluate this integral, it gives the precise area enclosed between these curves over the specified interval, capturing how the curves relate to each other spatially.