Question

The average value of a function \(f\left( {x,y} \right)\) over a rectangle \(R\) is defined to be .

Find the average value of \(f\) over the given rectangle, \(f\left( {x,y} \right) = {e^y}\sqrt {x + {e^y}} \), \(R = \left( {0,4} \right) \times \left( {0,1} \right)\).

Short answer

Given: \(f\left( {x,y} \right) = {e^y}\sqrt {x + {e^y}} \)

\(R = \left( {0,4} \right) \times \left( {0,1} \right)\)

To find: average value of\(f\).

Step-by-step solution

Using Integration, with respect to \(y\).

\({f_{ave}} = \frac{1}{4}\int\limits_0^4 {\int\limits_0^1 {{e^y}\sqrt {x + {e^y}} dydx} } \)

Integrate on \(y\). Use the substitution.

\(u = x + {e^y}\)and \(du = {e^y}dy\)

\begin{aligned} & {{f}_{ave}}=\frac{1}{4}\int\limits_{0}^{4}{\int\limits_{x+1}^{x+e}{{{u}^{{1}/{2}\;}}dxdx}} \\ & =\frac{1}{4}\int\limits_{0}^{0}{\frac{2}{3}\left( {{u}^{{3}/{2}\;}} \right)_{x+1}^{x+e}dx} \\ & =\frac{1}{6}\int\limits_{0}^{0}{{{\left( x+e \right)}^{{3}/{2}\;}}-{{\left( x+1 \right)}^{{3}/{2}\;}}dx} \end{aligned}

Integrating with respect to \(x\).

Note that for any Constant \(c\), if

\(\int {f\left( x \right)dx = F\left( x \right)} \)then,

\(\int {f\left( {x + c} \right)dx = F\left( {x + c} \right)} \)

\begin{aligned}{f_{ave}} &= \frac{1}{6}\left( {\frac{2}{5}{{\left( {x + e} \right)}^{{5 \mathord{\left/{\vphantom {5 2}} \right.} 2}}} - \frac{2}{5}{{\left( {x + 1} \right)}^{{5 \mathord{\left/{\vphantom {5 2}} \right.} 2}}}} \right)_0^4 \\ &= \frac{1}{{15}}\left[ {{{\left( {4 + e} \right)}^{{5 \mathord{\left/{\vphantom {5 2}} \right.} 2}}} - {{\left( e \right)}^{{5 \mathord{\left/{\vphantom {5 2}} \right.} 2}}} - {{\left( 5 \right)}^{{5 \mathord{\left/{\vphantom {5 2}} \right.} 2}}} + {{\left( 1 \right)}^{{5 \mathord{\left/{\vphantom {5 2}}\right.} 2}}}} \right] \\ &= \frac{1}{{15}}\left[ {{{\left( {4 + e} \right)}^{{5 \mathord{\left/{\vphantom {5 2}} \right.} 2}}} - {{\left( e \right)}^{{5 \mathord{\left/{\vphantom {5 2}} \right.} 2}}} - {{\left( 5 \right)}^{{5 \mathord{\left/{\vphantom {5 2}} \right.} 2}}} + 1} \right] \\ \end{aligned}

Therefore, average value of f is \(=\frac{1}{15}\left( {{\left( 4+e \right)}^{{5}/{2}\;}}-{{e}^{{5}/{2}\;}}-{{5}^{{5}/{2}\;}}+1 \right)\)