let f be the function defined by f(x) = sin(π/6 x). what is the average value of f on the interval 8, 10…

Explanation:

Step1: Recall average - value formula

The average value of a function $y = f(x)$ on the interval $[a,b]$ is given by $\bar{y}=\frac{1}{b - a}\int_{a}^{b}f(x)dx$. Here, $a = 8$, $b = 10$, and $f(x)=\sin(\frac{\pi}{6}x)$. So, $\bar{y}=\frac{1}{10 - 8}\int_{8}^{10}\sin(\frac{\pi}{6}x)dx=\frac{1}{2}\int_{8}^{10}\sin(\frac{\pi}{6}x)dx$.

Step2: Use substitution

Let $u=\frac{\pi}{6}x$, then $du=\frac{\pi}{6}dx$ and $dx=\frac{6}{\pi}du$. When $x = 8$, $u=\frac{\pi}{6}\times8=\frac{4\pi}{3}$; when $x = 10$, $u=\frac{\pi}{6}\times10=\frac{5\pi}{3}$. So, $\frac{1}{2}\int_{8}^{10}\sin(\frac{\pi}{6}x)dx=\frac{1}{2}\times\frac{6}{\pi}\int_{\frac{4\pi}{3}}^{\frac{5\pi}{3}}\sin(u)du=\frac{3}{\pi}\int_{\frac{4\pi}{3}}^{\frac{5\pi}{3}}\sin(u)du$.

Step3: Integrate $\sin(u)$

The antiderivative of $\sin(u)$ is $-\cos(u)$. So, $\frac{3}{\pi}[-\cos(u)]_{\frac{4\pi}{3}}^{\frac{5\pi}{3}}=\frac{3}{\pi}[-\cos(\frac{5\pi}{3})+\cos(\frac{4\pi}{3})]$.

Step4: Evaluate cosine values

We know that $\cos(\frac{5\pi}{3})=\frac{1}{2}$ and $\cos(\frac{4\pi}{3})=-\frac{1}{2}$. Then $\frac{3}{\pi}[-\frac{1}{2}+(-\frac{1}{2})]=\frac{3}{\pi}(-1)=-\frac{3}{\pi}$.

Answer:

$-\frac{3}{\pi}$