Question

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

Explanation:

Step1: Determine the upper - lower function

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 $f(x)$ and $g(x)$ at a value in the interval, say $x = 0$. $f(0)=2\cos(0)=2$ and $g(0)=2\sin(0)=0$. So, on the interval $[0,0.3\pi]$, $2\cos(x)\geq2\sin(x)$.

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

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

\]

Step4: Evaluate the definite integral

We know that $\sin(0) = 0$, $\cos(0)=1$, $\sin(0.3\pi)\approx0.809$ and $\cos(0.3\pi)\approx0.588$.
\[

$$\begin{align*} &2\times0.809 + 2\times0.588-2\times0 - 2\times1\\ =&1.618+1.176 - 2\\ =&0.794 \end{align*}$$

\]

Answer:

$0.794$