find the area of the shaded region shown in the graph.\nthe area of the shaded region is . (type an exact…

Explanation:

Step1: Find intersection point

Set $4\sin x = 4\cos x$, then $\tan x=1$, so $x = \frac{\pi}{4}$ in the given domain.

Step2: Set up integral for area

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $a$ to $b$ is $A=\int_{a}^{b}|f(x)-g(x)|dx$. Here, for $x\in[0,\frac{\pi}{4}]$, $4\cos x\geq4\sin x$ and for $x\in[\frac{\pi}{4},\frac{\pi}{2}]$, $4\sin x\geq4\cos x$. So $A=\int_{0}^{\frac{\pi}{4}}(4\cos x - 4\sin x)dx+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(4\sin x - 4\cos x)dx$.

Step3: Integrate first - part

$\int_{0}^{\frac{\pi}{4}}(4\cos x - 4\sin x)dx=4\int_{0}^{\frac{\pi}{4}}(\cos x-\sin x)dx=4[\sin x+\cos x]_0^{\frac{\pi}{4}}=4(\sin\frac{\pi}{4}+\cos\frac{\pi}{4}-\sin0 - \cos0)=4(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-0 - 1)=4(\sqrt{2}-1)$.

Step4: Integrate second - part

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(4\sin x - 4\cos x)dx=4\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin x-\cos x)dx=4[-\cos x-\sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}}=4(-\cos\frac{\pi}{2}-\sin\frac{\pi}{2}+\cos\frac{\pi}{4}+\sin\frac{\pi}{4})=4(0 - 1+\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2})=4(\sqrt{2}-1)$.

Step5: Sum up the two results

$A = 4(\sqrt{2}-1)+4(\sqrt{2}-1)=8(\sqrt{2}-1)$.

Answer:

$8(\sqrt{2}-1)$