Problem 36 - 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 below the line \(y=2\) and above the curve \(y=\sec ^{2} x\) on the interval \([0, \pi / 4]\)

Step-by-Step Solution

Answer

Answer: \(\frac{\pi}{2}-1\)

Step 1: Sketch the bounding curves and the region

To get an idea of the area we are trying to find, let's sketch the line \(y = 2\), the curve \(y = \sec^2 x\), and the interval \([0, \pi / 4]\) on the \(xy\)-plane. You will notice that the area we are trying to find lies below the line and above the curve, enclosed between the lines \(x=0\) and \(x=\pi/4\).

Step 2: Set up the integral

To find the area of the bounded region, we can integrate the difference between the two functions over the specified interval. The integral can be set up as follows: $$ \int_{0}^{\frac{\pi}{4}} (2 - \sec^2x) dx $$

Step 3: Evaluate the integral

Now let's proceed to evaluate the integral: $$ \int_{0}^{\frac{\pi}{4}} (2 - \sec^2x) dx = \int_{0}^{\frac{\pi}{4}} (2dx - \sec^2x\, dx) $$ We can evaluate the two separate integrals for \(2dx\) and \(\sec^2x\, dx\).First, let's evaluate the integral for \(2dx\): $$ \int_0^{\frac{\pi}{4}} 2 dx = 2x\Big|_0^{\frac{\pi}{4}} = 2\frac{\pi}{4} = \frac{\pi}{2} $$ Next, let's evaluate the integral for \(\sec^2 x \, dx\): $$ \int_0^{\frac{\pi}{4}} \sec^2 x \, dx = \tan x\Big|_0^{\frac{\pi}{4}} = 1-0 = 1 $$ Now, we can combine the two results: $$ \frac{\pi}{2}-1 $$

Step 4: Calculate the area

Using the result from the integral, the area of the region bounded by the line \(y = 2\), the curve \(y = \sec^2x\), and the interval \([0, \pi / 4]\) is: $$ Area = \frac{\pi}{2}-1 $$