Questions: Question 10, 5.6.91 Find the area of the region enclosed by the following curves y = sec^2 x, y = tan^2 x, x = -π/6, and x = π/6 The area of the region enclosed by the curves is . (Type an exact answer, using π as needed.)

Question 10, 5.6.91

Find the area of the region enclosed by the following curves
y = sec^2 x, y = tan^2 x, x = -π/6, and x = π/6

The area of the region enclosed by the curves is .
(Type an exact answer, using π as needed.)

Transcript text: Question 10, 5.6.91 Find the area of the region enclosed by the following curves $y=\sec ^{2} x, y=\tan ^{2} x, \quad x=-\frac{\pi}{6}$, and $x=\frac{\pi}{6}$ The area of the region enclosed by the curves is $\square$ . (Type an exact answer, using $\pi$ as needed.)

Solution

Solution Steps

To find the area of the region enclosed by the curves $ y = \sec^2 x $ and $ y = \tan^2 x $ between $ x = -\frac{\pi}{6} $ and $ x = \frac{\pi}{6} $, we need to set up an integral. The area can be found by integrating the difference of the two functions over the given interval.

Identify the interval of integration: $ x = -\frac{\pi}{6} $ to $ x = \frac{\pi}{6} $.
Determine the integrand: $ \sec^2 x - \tan^2 x $.
Integrate the function over the specified interval.

Step 1: Identify the Interval of Integration

We need to find the area enclosed by the curves $ y = \sec^2 x $ and $ y = \tan^2 x $ between $ x = -\frac{\pi}{6} $ and $ x = \frac{\pi}{6} $.

Step 2: Determine the Integrand

The integrand is the difference between the two functions: \[ \sec^2 x - \tan^2 x \]

Step 3: Set Up the Integral

We set up the integral of the difference of the functions over the interval from $ x = -\frac{\pi}{6} $ to $ x = \frac{\pi}{6} $: \[ \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} (\sec^2 x - \tan^2 x) \, dx \]

Step 4: Evaluate the Integral

Evaluating the integral, we get: \[ \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} (\sec^2 x - \tan^2 x) \, dx = \frac{\pi}{3} \]