/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 87 Find the area of the region. Use... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

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

Step by step solution

01

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.
02

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: \( \tan( \pi/3) - \tan( \pi/4) \).
03

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: \( \sqrt{3} - 1 \).
04

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 \( \sqrt{3} - 1 \).

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