Question

one - a - day multi vitamins
5 - 4 - 3 - 2 - 1
5 multiple choice: do one - a - day!

1. the average value of $\cos x$ on the interval $-3,5$ is

(a) $\frac{\sin 5-\sin 3}{8}$
(b) $\frac{\sin 5-\sin 3}{2}$
(c) $\frac{\sin 3-\sin 5}{2}$
(d) $\frac{\sin 3+\sin 5}{2}$
(e) $\frac{\sin 3+\sin 5}{8}$

Explanation:

Step1: Recall the 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, \(f(x)=\cos x\), \(a=-3\), and \(b = 5\). So the average value is \(\frac{1}{5-(-3)}\int_{-3}^{5}\cos xdx=\frac{1}{8}\int_{-3}^{5}\cos xdx\).

Step2: Integrate \(\cos x\)

We know that \(\int\cos xdx=\sin x + C\). Then \(\frac{1}{8}\int_{-3}^{5}\cos xdx=\frac{1}{8}[\sin x]_{-3}^{5}\).

Step3: Evaluate the definite - integral

Using the fundamental theorem of calculus \([\sin x]_{-3}^{5}=\sin(5)-\sin(-3)\). Since \(\sin(-x)=-\sin x\), then \(\sin(5)-\sin(-3)=\sin(5)+\sin(3)\).

Answer:

E. \(\frac{\sin3+\sin5}{8}\)