Question

(1 point) find the area of the region enclosed between (y = 2sin(x)) and (y = 2cos(x)) from (x = 0) to (x = 0.7pi). hint: notice that this region consists of two parts.

Explanation:

Step1: Determine the upper - lower functions

We need to find which function is on top and which is on the bottom in the interval \([0,0.3\pi]\). Let \(f(x)=2\cos(x)\) and \(g(x)=2\sin(x)\). Evaluate \(f(x)\) and \(g(x)\) at \(x = 0\): \(f(0)=2\cos(0)=2\) and \(g(0)=2\sin(0)=0\). Evaluate at \(x = 0.3\pi\): \(f(0.3\pi)=2\cos(0.3\pi)\approx2\times0.809 = 1.618\), \(g(0.3\pi)=2\sin(0.3\pi)\approx2\times0.588 = 1.176\). So \(y = 2\cos(x)\) is the upper - function and \(y = 2\sin(x)\) is the lower - function in the interval \([0,0.3\pi]\).

Step2: Use the area formula

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 \(\int_{0}^{0.3\pi}(2\cos(x)-2\sin(x))dx=2\int_{0}^{0.3\pi}\cos(x)dx-2\int_{0}^{0.3\pi}\sin(x)dx\).
\[

$$\begin{align*} 2\int_{0}^{0.3\pi}\cos(x)dx-2\int_{0}^{0.3\pi}\sin(x)dx&=2[\sin(x)]_{0}^{0.3\pi}+2[\cos(x)]_{0}^{0.3\pi}\\ &=2\sin(0.3\pi)+2\cos(0.3\pi)-2\sin(0)-2\cos(0)\\ &=2\times0.588 + 2\times0.809-0 - 2\\ &=1.176+1.618 - 2\\ &=0.794 \end{align*}$$

\]

Answer:

\(0.794\)