Problem 34 - Calculus Early Transcendentals

Question

Use any method (including geometry) to find the area of the following regions. In each case, sketch the bounding curves and the region in question. The region in the first quadrant bounded by \(y=2\) and \(y=2 \sin x\) on the interval \([0, \pi / 2]\)

Step-by-Step Solution

Answer

Answer: The area of the region is \(\pi - 2\) square units.

Step 1: Sketch the graphs

To visualize the bounded region, let's sketch the graphs of y=2 and y=2sin(x) for x in the interval [0, π/2]. 1. The graph of y=2 is a horizontal line passing through the point (0, 2). 2. The graph of y=2sin(x) is a sine wave with a maximum value of 2 at x=π/2. The bounded region is between the curves y=2 and y=2sin(x), and is enclosed in the first quadrant.

Step 2: Set Up the Integral

To find the area of the region, we can use integration. We will integrate the difference between the two functions (y=2 - y=2sin(x)) with respect to x, from x=0 to x=π/2. The integral we need to find is: \(\int_{0}^{\pi/2} (2 - 2 \sin x) \, dx\)

Step 3: Evaluate the Integral

Now we'll evaluate the integral: \(\int_{0}^{\pi/2} (2 - 2 \sin x) \, dx = \int_{0}^{\pi/2} 2 \, dx - \int_{0}^{\pi/2} 2 \sin x \, dx\) For the first integral: \(\int_{0}^{\pi/2} 2 \, dx = 2x\Big|_0^{\pi/2} = 2(\pi/2) - 2(0) = \pi\) For the second integral: \(\int_{0}^{\pi/2} 2 \sin x \, dx = -2\cos x\Big|_0^{\pi/2} = -2(\cos(\pi/2) - \cos(0)) = 2\) Now, subtract the results of the two integrals: \(\pi - 2 = \pi - 2\)

Step 4: Find the Area of the Region

The area of the region bounded by the curves y=2 and y=2sin(x) in the first quadrant and on the interval [0, π/2] is: Area = \(\pi - 2\) Thus, the area of the region is \(\pi - 2\) square units.