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Chapter 12: Problem 1087

The area of the region bounded by curves \(f(x)=\sin x\) \(g(x)=\cos x, x=(\pi / 4), x=(5 \pi / 4)\) is........ (a) 1 (b) 2 (c) \(\sqrt{2}\) (d) \(2 \sqrt{2}\)

Short Answer

Expert verified
The area of the region bounded by the curves \(f(x)=\sin x\), \(g(x)=\cos x\), \(x=(\pi / 4)\), and \(x=(5 \pi / 4)\) is (d) \(2 \sqrt{2}\).

Step by step solution

01

Determine which function is greater in the given interval

First, let's find where the two functions are equal by solving the following equation: \(f(x)=g(x)\) \[ \sin x = \cos x \] We can rewrite this equation as: \[ \tan x = 1 \] We know that \(\tan x\) will be equal to 1 at \(x = \pi/4\) (and the series \((2n + 1)\pi/4\), where \(n\) is an integer). Now let's check the given interval. Since \(x = (\pi/4)\) is one of the endpoints and \(\tan x = 1\) at that point, we need to check within the interval. For \(x=(\pi/2)\), which falls between the given endpoints, we get \(f(x) = \sin(\pi/2) = 1\) and \(g(x) = \cos(\pi/2) = 0\). Therefore, \(f(x) > g(x)\) within the interval.
02

Integrate the difference within the interval

Since \(f(x) > g(x)\) in the interval, we will integrate the difference \((\sin x - \cos x)\) between the two functions. Let \(A\) be the area of the region: \[ A = \int_{\pi/4}^{5\pi/4} (\sin x - \cos x) dx \]
03

Integrate and substitute the limits of integration

Now integrate the functions separately and evaluate the definite integral: \[ A = \left[-\cos x - \sin x\right]_{\pi/4}^{5\pi/4} \] Now, substitute the limits of integration: \[ A = \left[(-\cos(5\pi/4)-\sin(5\pi/4))\) - \((-cos(\pi/4)-\sin(\pi/4))\right] \] By simplifying the trigonometric expressions, we get: \[ A = [(\sqrt{2}/2-\sqrt{2}/2)]-[(-\sqrt{2}/2+\sqrt{2}/2)] \] \[ A = 2(\sqrt{2}/2 - (-\sqrt{2}/2)) \] \[ A = 2(\sqrt{2}) \] Therefore, the answer is (d) \(2\sqrt{2}\).

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