what is the average value of sin(x) on the interval -6 ≤ x ≤ -1? choose 1 answer: a (cos(6) + cos(1))/5 b…

Explanation:

Step1: Recall average - value formula

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

Step2: Integrate $\sin(x)$

We know that $\int\sin(x)dx=-\cos(x)+C$. Then $\frac{1}{5}\int_{-6}^{-1}\sin(x)dx=\frac{1}{5}[-\cos(x)]_{-6}^{-1}$.

Step3: Evaluate the definite - integral

Using the fundamental theorem of calculus $[-\cos(x)]_{-6}^{-1}=-\cos(-1)+\cos(-6)$. Since $\cos(-\theta)=\cos(\theta)$ for any real number $\theta$, we have $-\cos(-1)+\cos(-6)=\cos(6)-\cos(1)$. So the average value is $\frac{\cos(6)-\cos(1)}{5}$.

Answer:

B. $\frac{\cos(6)-\cos(1)}{5}$