Question

homework5: problem 9 (1 point) find the area of the region enclosed between y = 2 sin(x) and y = 2 cos(x) from x = 0 to x = 0.3π. hint: notice that this region consists of two parts. you have attempted this problem 0 times. you have unlimited attempts remaining. previous problem problem list next problem preview my answers submit answers

Explanation:

Step1: Determine the upper - lower functions

We need to find which function is greater on the interval \([0,0.3\pi]\). Let \(f(x)=2\cos(x)\) and \(g(x)=2\sin(x)\). Evaluate \(f(x)-g(x)=2\cos(x)-2\sin(x)\) at \(x = 0\), \(f(0)-g(0)=2\cos(0)-2\sin(0)=2>0\), so \(y = 2\cos(x)\) is above \(y = 2\sin(x)\) on the interval \([0,0.3\pi]\).

Step2: Use the area formula for the region between two curves

The area \(A\) between two curves \(y = f(x)\) and \(y = g(x)\) from \(x=a\) to \(x = b\) is given by \(A=\int_{a}^{b}[f(x)-g(x)]dx\). Here, \(a = 0\), \(b = 0.3\pi\), \(f(x)=2\cos(x)\) and \(g(x)=2\sin(x)\), so \(A=\int_{0}^{0.3\pi}(2\cos(x)-2\sin(x))dx\).

Step3: Integrate term - by - term

We know that \(\int\cos(x)dx=\sin(x)+C\) and \(\int\sin(x)dx=-\cos(x)+C\). Then \(\int_{0}^{0.3\pi}(2\cos(x)-2\sin(x))dx=2\int_{0}^{0.3\pi}\cos(x)dx-2\int_{0}^{0.3\pi}\sin(x)dx\).
\[

$$\begin{align*} 2\int_{0}^{0.3\pi}\cos(x)dx-2\int_{0}^{0.3\pi}\sin(x)dx&=2[\sin(x)]_{0}^{0.3\pi}+2[\cos(x)]_{0}^{0.3\pi}\\ &=2\sin(0.3\pi)+2\cos(0.3\pi)-2\sin(0)-2\cos(0)\\ &=2\sin(0.3\pi)+2\cos(0.3\pi)- 2 \end{align*}$$

\]
Using \(\sin(0.3\pi)\approx0.809\) and \(\cos(0.3\pi)\approx0.588\), we have \(A = 2\times0.809+2\times0.588 - 2=1.618 + 1.176-2=0.794\)

Answer:

\(2\sin(0.3\pi)+2\cos(0.3\pi)-2\approx0.794\)