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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 6 - Applications of IntegrationProblema 6.2.65b
Capitolo 6, Problema 6.2.65b

Determine whether the following statements are true and give an explanation or counterexample.


b. The area of the region between y=sin x and y=cos x on the interval [0,π/2] is ∫π/20(cosx−sinx)dx.

Guida verificata passo dopo passo
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First, identify the two functions given: \(y = \sin x\) and \(y = \cos x\) on the interval \([0, \frac{\pi}{2}]\).
Determine which function is on top (greater) and which is on the bottom over the interval \([0, \frac{\pi}{2}]\). This is important because the area between two curves is found by integrating the difference between the top function and the bottom function.
Evaluate the values of \(\sin x\) and \(\cos x\) at the endpoints: at \(x=0\), \(\sin 0 = 0\) and \(\cos 0 = 1\), so \(\cos x\) is greater; at \(x=\frac{\pi}{2}\), \(\sin \frac{\pi}{2} = 1\) and \(\cos \frac{\pi}{2} = 0\), so \(\sin x\) is greater. This means the two curves cross somewhere in the interval.
Find the point where \(\sin x = \cos x\) in \([0, \frac{\pi}{2}]\) by solving \(\sin x = \cos x\). This will give the exact point where the top and bottom functions switch.
To find the total area between the curves, split the integral at the crossing point and integrate the absolute difference accordingly: \(\int_0^{c} (\cos x - \sin x) \, dx + \int_c^{\frac{\pi}{2}} (\sin x - \cos x) \, dx\), where \(c\) is the crossing point. Therefore, the statement that the area is \(\int_0^{\frac{\pi}{2}} (\cos x - \sin x) \, dx\) without splitting is not correct.

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