Chapter 6.3, Problem 49E

To determine

To choose: The correct option for the given condition.

Answer to Problem 49E

Correct option is (B).

Explanation of Solution

Given information :

Cosine function on the interval [1,5]

  $\begin{array}{cc}av(\cos x)& =\frac{1}{5-1}{{\displaystyle \int }}_1^5\cos xdx\\ & =\frac{1}{4}{{\displaystyle \int }}_1^5\cos xdx\end{array}$

Calculation :

If $\text{f}$ is integrated on $[a,b],$ , its average (mean) value on $[a,b]$ is

  $av(f)=\frac{1}{b-a}{{\displaystyle \int }}_a^{b}f(x)dx$

So,

  $\begin{array}{cc}av(\cos x)& =\frac{1}{5-1}{{\displaystyle \int }}_1^5\cos xdx\\ & =\frac{1}{4}{{\displaystyle \int }}_1^5\cos xdx\end{array}$

Now calculate the integral-

Since $\cos x$ is the rate of change of the quantity

  $F(x)=\sin x$

That is,

  $F^{\prime }(x)=\cos x$

Since,

  ${{\displaystyle \int }}_a^{x}f(t)dt=F(x)-F(a)$

Where F is any anti-derivative of f

So,

  $\begin{array}{cc}{{\displaystyle \int }}_1^5\cos xdx& =\sin (5)-(\sin 1)\\ & \approx (-0.959)-(0.841)\\ & \approx -1.8\end{array}$

So,

  $\begin{array}{cc}av(\cos x)& \approx \frac{1}{4}(-1.8)\\ & \approx -0.450\end{array}$