Question

(1 point) find the area of the region enclosed between (y = 2sin(x)) and (y = 2cos(x)) from (x = 0) to (x = 0.3pi). hint: notice that the area is not correct. your score was recorded. your score was successfully sent to the lms. you have attempted this problem 1 time. you received a score of 0% for this attempt. your overall recorded score is 0%. you have unlimited attempts remaining.

Explanation:

Step1: Find intersection points

Set \(2\sin(x)=2\cos(x)\), then \(\tan(x) = 1\), \(x=\frac{\pi}{4}+k\pi,k\in\mathbb{Z}\). In the interval \([0, 0.3\pi]\), the intersection - point is \(x = \frac{\pi}{4}\).

Step2: Determine which function is on top

For \(x\in[0,\frac{\pi}{4}]\), \(2\cos(x)\geq2\sin(x)\); for \(x\in[\frac{\pi}{4},0.3\pi]\), \(2\sin(x)\geq2\cos(x)\).

Step3: Use the area formula for the region between two curves

The area \(A=\int_{0}^{\frac{\pi}{4}}(2\cos(x)-2\sin(x))dx+\int_{\frac{\pi}{4}}^{0.3\pi}(2\sin(x)-2\cos(x))dx\).
First integral: \(\int_{0}^{\frac{\pi}{4}}(2\cos(x)-2\sin(x))dx=2\int_{0}^{\frac{\pi}{4}}(\cos(x)-\sin(x))dx=2[\sin(x)+\cos(x)]_{0}^{\frac{\pi}{4}}=2(\sin\frac{\pi}{4}+\cos\frac{\pi}{4}-\sin(0)-\cos(0))=2(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-0 - 1)=2(\sqrt{2}-1)\).
Second integral: \(\int_{\frac{\pi}{4}}^{0.3\pi}(2\sin(x)-2\cos(x))dx=2\int_{\frac{\pi}{4}}^{0.3\pi}(\sin(x)-\cos(x))dx=2[-\cos(x)-\sin(x)]_{\frac{\pi}{4}}^{0.3\pi}=2(-\cos(0.3\pi)-\sin(0.3\pi)+\cos\frac{\pi}{4}+\sin\frac{\pi}{4})\).
\(\cos(0.3\pi)\approx0.809\), \(\sin(0.3\pi)\approx0.588\), \(\cos\frac{\pi}{4}=\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}\approx0.707\).
\(A = 2(\sqrt{2}-1)+2(-\cos(0.3\pi)-\sin(0.3\pi)+\cos\frac{\pi}{4}+\sin\frac{\pi}{4})\)
\(A=2\sqrt{2}-2+2(- 0.809 - 0.588+0.707 + 0.707)\)
\(A=2\sqrt{2}-2+2(-1.397 + 1.414)\)
\(A=2\sqrt{2}-2 + 2\times0.017\)
\(A=2\sqrt{2}-2 + 0.034\)
\(A\approx2\times1.414-2 + 0.034\)
\(A\approx2.828-2 + 0.034\)
\(A\approx0.862\)

Answer:

\(0.862\)