Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.2 Exercises - Page 523: 87

${f_{avg}} \approx 0.223$

$$\eqalign{ & {\text{Let }}y = {e^{ - 4t}}\left( {\cos 2t + 5\sin 2t} \right) \cr & {\text{The average value on the interval }}\left[ {0,\pi } \right]{\text{ is given by}} \cr & {f_{avg}} = \frac{1}{{\pi - 0}}\int_0^\pi {{e^{ - 4t}}\left( {\cos 2t + 5\sin 2t} \right)} dt \cr & {f_{avg}} = \frac{1}{\pi }\int_0^\pi {{e^{ - 4t}}\left( {\cos 2t + 5\sin 2t} \right)} dt \cr & {f_{avg}} = \frac{1}{\pi }\int_0^\pi {\left( {{e^{ - 4t}}\cos 2t + 5{e^{ - 4t}}\sin 2t} \right)} dt \cr & \cr & *{\text{Integrating }}\int {{e^{ - 4t}}\cos 2t} dt{\text{ using the formula of exercise 71}} \cr & {\text{Let }}u = {e^{ - 4t}},{\text{ }}du = - 4{e^{ - 4t}}dt \cr & dv = \cos 2t,{\text{ }}v = \frac{1}{2}\sin 2t \cr & {\text{Using integration by parts formula}} \cr & \int {{e^{ - 4t}}\cos 2t} dt = \frac{1}{2}{e^{ - 4t}}\sin 2t + \int {\frac{1}{2}\sin 2t\left( {4{e^{ - 4t}}} \right)dt} \cr & \int {{e^{ - 4t}}\cos 2t} dt = \frac{1}{2}{e^{ - 4t}}\sin 2t + \int {{e^{ - 4t}}\sin 2t\left( 2 \right)dt} \cr & {\text{Let }}u = {e^{ - 4t}},{\text{ }}du = - 4{e^{ - 4t}}dt \cr & dv = \sin 2t\left( 2 \right),{\text{ }}v = - \cos 2t \cr & \int {{e^{ - 4t}}\cos 2t} dt = \frac{1}{2}{e^{ - 4t}}\sin 2t - {e^{ - 4t}}\cos 2t - \int {\left( { - \cos 2t} \right)} \left( { - 4{e^{ - 4t}}} \right)dt \cr & \int {{e^{ - 4t}}\cos 2t} dt = \frac{1}{2}{e^{ - 4t}}\sin 2t - {e^{ - 4t}}\cos 2t - 4\int {{e^{ - 4t}}\cos 2t} dt \cr & {\text{Solve for }}\int {{e^{ - 4t}}\cos 2t} dt \cr & 5\int {{e^{ - 4t}}\cos 2t} dt = \frac{1}{2}{e^{ - 4t}}\sin 2t - {e^{ - 4t}}\cos 2t \cr & \int {{e^{ - 4t}}\cos 2t} dt = \frac{1}{{10}}{e^{ - 4t}}\sin 2t - \frac{1}{5}{e^{ - 4t}}\cos 2t + C \cr & \cr & *{\text{Integrating }}\int {5{e^{ - 4t}}\sin 2t} dt{\text{ by parts}} \cr & {\text{Let }}u = 5{e^{ - 4t}},{\text{ }}du = - 20{e^{ - 4t}}dt \cr & dv = \sin 2t,{\text{ }}v = - \frac{1}{2}\cos 2t \cr & {\text{Using integration by parts formula}} \cr & \int {5{e^{ - 4t}}\sin 2t} dt = - \frac{5}{2}{e^{ - 4t}}\cos 2t - \int {\left( { - \frac{1}{2}\cos 2t} \right)\left( { - 20{e^{ - 4t}}} \right)dt} \cr & \int {5{e^{ - 4t}}\sin 2t} dt = - \frac{5}{2}{e^{ - 4t}}\cos 2t - \int {5{e^{ - 4t}}\cos 2t\left( 2 \right)dt} \cr & {\text{Let }}u = 5{e^{ - 4t}},{\text{ }}du = - 20{e^{ - 4t}}dt \cr & dv = \cos 2t\left( 2 \right),{\text{ }}v = \sin 2t \cr & \int {5{e^{ - 4t}}\sin 2t} dt = - \frac{5}{2}{e^{ - 4t}}\cos 2t - 5{e^{ - 4t}}\sin 2t + \int {\sin 2t} \left( { - 20{e^{ - 4t}}} \right)dt \cr & \int {5{e^{ - 4t}}\sin 2t} dt = - \frac{5}{2}{e^{ - 4t}}\cos 2t - 5{e^{ - 4t}}\sin 2t - 20\int {{e^{ - 4t}}\sin 2t} dt \cr & {\text{Solve for }}\int {{e^{ - 4t}}\sin 2t} dt \cr & \int {{e^{ - 4t}}\sin 2t} dt = - {e^{ - 4t}}\sin 2t - \frac{1}{2}{e^{ - 4t}}\cos 2t + C \cr & {\text{Therefore}}{\text{,}} \cr & {f_{avg}} = \frac{1}{\pi }\left[ {\frac{1}{{10}}{e^{ - 4t}}\sin 2t - \frac{1}{5}{e^{ - 4t}}\cos 2t} \right]_0^\pi \cr & + \frac{1}{\pi }\left[ { - {e^{ - 4t}}\sin 2t - \frac{1}{2}{e^{ - 4t}}\cos 2t} \right]_0^\pi \cr & {f_{avg}} = \frac{1}{\pi }\left[ { - \frac{9}{{10}}{e^{ - 4t}}\sin 2t - \frac{7}{{10}}{e^{ - 4t}}\cos 2t} \right]_0^\pi \cr & {\text{Evaluating the limits}} \cr & {f_{avg}} = \frac{1}{\pi }\left[ { - \frac{9}{{10}}{e^{ - 4\pi }}\sin 2\pi - \frac{7}{{10}}{e^{ - 4\pi }}\cos 2\pi } \right] \cr & - \frac{1}{\pi }\left[ { - \frac{9}{{10}}{e^0}\sin 0 - \frac{7}{{10}}{e^0}\cos 0} \right] \cr & {\text{Simplify}} \cr & {f_{avg}} = \frac{1}{\pi }\left[ {0 - \frac{7}{{10}}{e^{ - 4\pi }}} \right] - \frac{1}{\pi }\left[ {0 - \frac{7}{{10}}} \right] \cr & {f_{avg}} = - \frac{7}{{10\pi }}{e^{ - 4\pi }} + \frac{7}{{10\pi }} \cr & {f_{avg}} = \frac{7}{{10\pi }}\left( {1 - {e^{ - 4\pi }}} \right) \cr & {f_{avg}} \approx 0.223 \cr} $$