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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 8, Problem 8.8.87

87. Find the area of the region that lies between the curves y = sec x and y = tan x from x = 0 to x = π/2.

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Identify the curves and the interval: We are given two functions, \(y = \sec x\) and \(y = \tan x\), and we want to find the area between them from \(x = 0\) to \(x = \frac{\pi}{2}\).
Determine which function is on top and which is on the bottom in the interval \([0, \frac{\pi}{2})\): Since \(\sec x = \frac{1}{\cos x}\) and \(\tan x = \frac{\sin x}{\cos x}\), and for \(x\) in this interval \(\sin x\) is between 0 and 1, it follows that \(\sec x \geq \tan x\). So, the area between the curves is given by the integral of \((\sec x - \tan x)\) over \([0, \frac{\pi}{2})\).
Set up the definite integral for the area: The area \(A\) is given by \(A = \int_0^{\frac{\pi}{2}} (\sec x - \tan x) \, dx\).
Recall the antiderivatives: The integral of \(\sec x\) is \(\ln |\sec x + \tan x| + C\), and the integral of \(\tan x\) is \(-\ln |\cos x| + C\). Use these to find the antiderivative of the integrand.
Evaluate the definite integral by substituting the limits \(x = 0\) and \(x = \frac{\pi}{2}\) into the antiderivative expression, and then compute the difference to find the area.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Definite Integrals for Area Between Curves

The area between two curves over an interval is found by integrating the difference of their functions. Specifically, the integral of the upper function minus the lower function from the lower to the upper limit gives the enclosed area.
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Understanding the behavior of sec x and tan x on the interval [0, π/2) is crucial. Both functions increase and approach infinity as x approaches π/2, with sec x always greater than tan x in this interval, which determines the order of subtraction in the integral.
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