Question

What is the average value of $y=\frac{\cos x}{x^2+x+2}$ on the closed interval $\left[-1,3\right]$?

Answer 1

Find the average value of $y=\frac{\cos x}{x^2+x+2}$:

Consider the given function as:

$y=\frac{\cos x}{x^2+x+2}$

And the given interval is $\left[-1,3\right]$

Step-1: Calculate the value of the function at points $x=-1$ and $x=3$

$$\begin{gathered} y_{-1}=\frac{\cos \left(-1\right)}{{\left(-1\right)}^2+\left(-1\right)+2} \\ {y}_{-1}=\frac{\cos \left(-1\right)}{1-1+2} \\ {y}_{-1}=\frac{0.9998}{2} \\ {y}_{-1}=0.4999 \end{gathered}$$

Step-2 : Calculate the value of the function at points $x=3$

$$\begin{gathered} y_3=\frac{\cos \left(3\right)}{{\left(3\right)}^2+\left(3\right)+2} \\ {y}_3=\frac{\cos \left(3\right)}{9+3+2} \\ {y}_3=\frac{0.9988}{14} \\ {y}_3=0.0713 \end{gathered}$$

Step-3 : Find the average value of the function.

$$\begin{gathered} Average=\frac{y_3+y_{-1}}{3-\left(-1\right)} \\ Average=\frac{y_3+y_{-1}}{3+1} \end{gathered}$$

Substitute $y_3=0.0713$ and $y_{-1}=0.4999$ in the above Equation becomes,

$$\begin{gathered} Average=\frac{0.0713+0.4999}{3+1} \\ Average=\frac{0.5712}{4} \\ Average=0.1428 \end{gathered}$$

Hence, the average value of the function on the closed interval is $0.1428$.