5. Let $R$ be the region in the first quadrant under the graph of $y = \frac { x } { x ^ { 2 } + 2 }$ for $0 \leqq x \leqq \sqrt { 6 }$.\\
(a) Find the area of $R$.\\
(b) If the line $\mathrm { x } = k$ divides $R$ into two regions of equal area, what is the value of $k$ ?\\
(c) What is the average value of $y = \frac { x } { x ^ { 2 } + 2 }$ on the interval $0 \leqq x \leqq \sqrt { 6 }$ ?

Solution\\
Distribution of Points\\
(a) $A = \int _ { 0 } ^ { \sqrt { 6 } } \frac { x } { x ^ { 2 } + 2 } d x$

$$\begin{aligned}
& = \left. \frac { 1 } { 2 } \ln \left( x ^ { 2 } + 2 \right) \right| _ { 0 } ^ { \sqrt { 6 } } \\
& = \frac { 1 } { 2 } \ln 8 - \frac { 1 } { 2 } \ln 2 = \ln 2
\end{aligned}$$

(b) $\frac { 1 } { 2 } \ln 2 = \int _ { 0 } ^ { k } \frac { x } { x ^ { 2 } + 2 } d x$

$$\begin{aligned}
& = \left. \frac { 1 } { 2 } \ln \left( x ^ { 2 } + 2 \right) \right| _ { 0 } ^ { k } \\
& = \frac { 1 } { 2 } \ln \left( k ^ { 2 } + 2 \right) - \frac { 1 } { 2 } \ln 2
\end{aligned}$$

$$\begin{aligned}
& \therefore \frac { 1 } { 2 } \ln \left( k ^ { 2 } + 2 \right) = \frac { 1 } { 2 } \ln 2 + \frac { 1 } { 2 } \ln 2 \\
& \quad \text { or } \ln \left( k ^ { 2 } + 2 \right) = \ln 4 \\
& \therefore k ^ { 2 } + 2 = 4 \text { and } k = \sqrt { 2 }
\end{aligned}$$

(c) Average value $= \frac { 1 } { \sqrt { 6 } - 0 } \int _ { 0 } ^ { \sqrt { 6 } } \frac { \mathrm { x } } { \mathrm { x } ^ { 2 } + 2 } \mathrm { dx }$

$$= \frac { 1 } { \sqrt { 6 } } \ln 2$$

(a) 1: for correct integral\\
$3 : \left\{ \begin{array} { l } 1 : \text { for antiderivative } \\ 1 : \text { for evaluation } \end{array} \right.$\\
(b) $\quad \begin{cases} 1 : & \text { for a correct equation } \\ & \text { involving integral(s) } \\ 1 : & \text { for antiderivative } \\ 1 : & \text { for evaluation of integral(s) } \end{cases}$

1: for finding the value of $k$\\
(c) $2 : \left\{ \begin{array} { l } 1 : \text { for correct integral } \\ 1 : \text { for evaluation } \end{array} \right.$\\