Answer
$${f_{avg}} = 793800$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = 1260 - 315\left( {{e^{0.00418x}} + {e^{ - 0.00418x}}} \right){\text{ on}}\left[ { - 315,315} \right] \cr
& {\text{Find }}{f_{avg}}\left( x \right) \cr
& {f_{avg}} = \frac{1}{{315 - \left( { - 315} \right)}}\int_{ - 315}^{315} {\left[ {1260 - 315\left( {{e^{0.00418x}} + {e^{ - 0.00418x}}} \right)} \right]} dx \cr
& {\text{Identify the symmetry of }}1260 - 315\left( {{e^{0.00418x}} + {e^{ - 0.00418x}}} \right){\text{ }} \cr
& {\text{evaluate }}f\left( { - x} \right) \cr
& f\left( { - x} \right) = 1260 - 315\left( {{e^{0.00418\left( { - x} \right)}} + {e^{ - 0.00418\left( { - x} \right)}}} \right){\text{ }} \cr
& f\left( { - x} \right) = 1260 - 315\left( {{e^{ - 0.00418\left( { - x} \right)}} + {e^{0.00418x}}} \right){\text{ }} \cr
& f\left( { - x} \right) = f\left( x \right) \cr
& {\text{Therefore }}1260 - 315\left( {{e^{ - 0.00418\left( { - x} \right)}} + {e^{0.00418x}}} \right){\text{ is an even function}}{\text{.}} \cr
& {\text{Use the integral property }}\int_{ - a}^a {f\left( x \right)} dx = 2\int_0^2 {f\left( x \right)} dx,{\text{ }}f{\text{ is even}} \cr
& {f_{avg}} = 2\int_0^{315} {\left[ {1260 - 315\left( {{e^{0.00418x}} + {e^{ - 0.00418x}}} \right)} \right]} dx \cr
& {\text{Integrating}} \cr
& {f_{avg}} = 2\left[ {1260x - 315\left( {\frac{1}{{0.00418}}{e^{0.00418x}} - \frac{1}{{0.00418}}{e^{ - 0.00418x}}} \right)} \right]_0^{315} \cr
& {f_{avg}} = 2\left[ {1260x - \frac{{315}}{{0.00418}}\left( {{e^{0.00418x}} - {e^{ - 0.00418x}}} \right)} \right]_0^{315} \cr
& {\text{Evaluating}} \cr
& {f_{avg}} = 2\left[ {1260\left( {315} \right) - \frac{{315}}{{0.00418}}\left( {{e^{0.00418\left( {315} \right)}} - {e^{ - 0.00418\left( {315} \right)}}} \right)} \right] \cr
& - 2\left[ {1260\left( 0 \right) - \frac{{315}}{{0.00418}}\left( {{e^{0.00418\left( 0 \right)}} - {e^{ - 0.00418\left( 0 \right)}}} \right)} \right] \cr
& {f_{avg}} = 793800 \cr} $$
