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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 8, Problema 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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