Question

Find the area of the region. Use a graphing utility to verify your result. $$ \int_{\pi / 2}^{2 \pi / 3} \sec ^{2}\left(\frac{x}{2}\right) d x $$

Short answer

The area of the region is \( \sqrt{3} - 1 \).

Step-by-step solution

Understanding the integral

The integral is \(\int_{\pi / 2}^{2 \pi / 3} \sec ^{2}\left(\frac{x}{2}\right) d x \). Integration will provide the area under the curve that is limited by the given integral point.

Performing Integration

The integral of \( \sec^2(z) \) with respect to \( z \) is \( \tan(z) \). We will use the chain rule, where \( z = x/2 \). Thus, when \( x = \pi / 2 \), then \( z = \pi / 4 \) and when \( x = 2\pi / 3 \), then \( z = \pi / 3 \). Applying the fundamental theorem of calculus we end with: \( \tan( \pi/3) - \tan( \pi/4) \).

Calculation

The tangent of \( \pi/3 \) is \( \sqrt{3} \) and the tangent of \( \pi/4 \) is \( 1 \). Thus we get the final result after calculation: \( \sqrt{3} - 1 \).

Verifying Results with Graphing Utility

Plot the function \( \sec^2(x/2) \) and use the graphing utility to calculate the area between the curve and the x-axis over the interval \( \pi/2 \) to \( 2\pi/3 \). The result should match the earlier calculation, that is \( \sqrt{3} - 1 \).