Chapter 14 - Section 14.5 - Improper Integrals and Applications - Exercises - Page 1059: 10

$$ - 2$$

$$\eqalign{ & \int_0^{ + \infty } {\left( {2x - 4} \right){e^{ - x}}dx} \cr & {\text{Using the definition of improper integrals see page 1053}} \cr & \underbrace {\int_a^{ + \infty } {f\left( x \right)dx = \mathop {\lim }\limits_{M \to + \infty } } \int_a^M {f\left( x \right)} dx}_ \Downarrow \cr & \int_0^{ + \infty } {\left( {2x - 4} \right){e^{ - x}}dx} = \mathop {\lim }\limits_{M \to + \infty } \int_0^M {\left( {2x - 4} \right){e^{ - x}}dx} \cr & {\text{Integrating by parts}} \cr & u = 2x - 4,{\text{ }}du = 2dx \cr & dv = {e^{ - x}}dx,{\text{ }}v = - {e^{ - x}} \cr & \int {udv} = uv - \int {vdu} \cr & = - \left( {2x - 4} \right){e^{ - x}} + \int {2{e^{ - x}}dx} \cr & = - \left( {2x - 4} \right){e^{ - x}} - 2{e^{ - x}} + C \cr & \mathop {\lim }\limits_{M \to + \infty } \int_0^M {\left( {2x - 4} \right){e^{ - x}}dx} \cr & = \mathop {\lim }\limits_{M \to + \infty } \left[ { - \left( {2x - 4} \right){e^{ - x}} - 2{e^{ - x}}} \right]_0^M \cr & = \mathop {\lim }\limits_{M \to + \infty } \left[ { - 2x{e^{ - x}} + 4{e^{ - x}} - 2{e^{ - x}}} \right]_0^M \cr & = \mathop {\lim }\limits_{M \to + \infty } \left[ { - 2x{e^{ - x}} + 2{e^{ - x}}} \right]_0^M \cr & = \mathop {\lim }\limits_{M \to + \infty } \left( { - 2M{e^{ - M}} + 2{e^{ - M}} - 2\left( 0 \right){e^{ - x}} - 2{e^0}} \right) \cr & = \mathop {\lim }\limits_{M \to + \infty } \left( { - 2M{e^{ - M}} + 2{e^{ - M}} - 2{e^0}} \right) \cr & {\text{Evaluate the limit when }}M \to + \infty \cr & = \left[ {\left( { - 2\left( \infty \right){e^{ - \infty }} + 6{e^{ - \infty }}} \right) - \left( 2 \right)} \right] \cr & = 0 + 0 - 2 \cr & = - 2 \cr} $$