c1 - average value of a function: problem\n(1 point)\nfind the average value of : ( f ( x ) = 8 sin x + 5…

Answer:

$\frac{3}{17\pi}\left(-8\cos\left(\frac{17\pi}{6}\right)+5\sin\left(\frac{17\pi}{6}\right)+8\right)$

Explanation:

Step1: Recall the formula for the average value of a function

The average value of a function (y = f(x)) on the interval ([a,b]) is given by (f_{avg}=\frac{1}{b - a}\int_{a}^{b}f(x)dx). Here, (a = 0), (b=\frac{17\pi}{6}), and (f(x)=8\sin x+5\cos x). So, (f_{avg}=\frac{1}{\frac{17\pi}{6}-0}\int_{0}^{\frac{17\pi}{6}}(8\sin x + 5\cos x)dx=\frac{6}{17\pi}\int_{0}^{\frac{17\pi}{6}}(8\sin x + 5\cos x)dx).

Step2: Integrate the function

We know that (\int\sin xdx=-\cos x + C) and (\int\cos xdx=\sin x + C). Then (\int(8\sin x + 5\cos x)dx=-8\cos x+5\sin x + C).

Step3: Evaluate the definite - integral

Using the fundamental theorem of calculus (\int_{0}^{\frac{17\pi}{6}}(8\sin x + 5\cos x)dx=\left[-8\cos x+5\sin x\right]{0}^{\frac{17\pi}{6}}). [ \begin{align*} &(-8\cos\left(\frac{17\pi}{6}\right)+5\sin\left(\frac{17\pi}{6}\right))-(-8\cos(0)+5\sin(0))\ =&-8\cos\left(\frac{17\pi}{6}\right)+5\sin\left(\frac{17\pi}{6}\right)+8 \end{align*} ] Since (\cos\left(\frac{17\pi}{6}\right)=\cos\left(2\pi+\frac{5\pi}{6}\right)=\cos\left(\frac{5\pi}{6}\right)=-\frac{\sqrt{3}}{2}) and (\sin\left(\frac{17\pi}{6}\right)=\sin\left(2\pi+\frac{5\pi}{6}\right)=\sin\left(\frac{5\pi}{6}\right)=\frac{1}{2}). [ \begin{align*} -8\cos\left(\frac{17\pi}{6}\right)+5\sin\left(\frac{17\pi}{6}\right)+8&=-8\times\left(-\frac{\sqrt{3}}{2}\right)+5\times\frac{1}{2}+8\ &=4\sqrt{3}+\frac{5 + 16}{2}\ &=4\sqrt{3}+\frac{21}{2} \end{align*} ] And (f{avg}=\frac{6}{17\pi}\left(-8\cos\left(\frac{17\pi}{6}\right)+5\sin\left(\frac{17\pi}{6}\right)+8\right)=\frac{3}{17\pi}\left(-8\cos\left(\frac{17\pi}{6}\right)+5\sin\left(\frac{17\pi}{6}\right)+8\right))