Provide an appropriate response.
-(a) Show that $\int _ { 2 } ^ { \infty } e ^ { - 2 x } \mathrm { dx } = \frac { 1 } { 2 } \mathrm { e } ^ { - 4 }

Question

Provide an appropriate response.
-(a) Show that $\int _ { 2 } ^ { \infty } e ^ { - 2 x } \mathrm { dx } = \frac { 1 } { 2 } \mathrm { e } ^ { - 4 } < 0.0092$ and hence that $\int _ { 2 } ^ { \infty } \mathrm { e } ^ { - \mathrm { x } ^ { 2 } } \mathrm { dx } < 0.0092$.
(b) Explain why this means that $\int _ { 0 } ^ { \infty } \mathrm { e } ^ { - \mathrm { x } ^ { 2 } } \mathrm { dx }$ can be replaced by $\int _ { 0 } ^ { 2 } \mathrm { e } ^ { - \mathrm { x } ^ { 2 } } \mathrm { dx }$ without introducing an error of magnitude greater than $0.0092 .$

Quizplus · Accepted Answer

The Answer of Provide an appropriate response.
-(a) Show that $\int _ { 2 } ^ { \infty } e ^ { - 2 x } \mathrm { dx } = \frac { 1 } { 2 } \mathrm { e } ^ { - 4 } < 0.0092$ and hence that $\int _ { 2 } ^ { \infty } \mathrm { e } ^ { - \mathrm { x } ^ { 2 } } \mathrm { dx } < 0.0092$.
(b) Explain why this means that $\int _ { 0 } ^ { \infty } \mathrm { e } ^ { - \mathrm { x } ^ { 2 } } \mathrm { dx }$ can be replaced by $\int _ { 0 } ^ { 2 } \mathrm { e } ^ { - \mathrm { x } ^ { 2 } } \mathrm { dx }$ without introducing an error of magnitude greater than $0.0092 .$

Provide an Appropriate Response ∫2∞e−2xdx=12e−4<0.0092\int _ { 2 } ^ { \infty } e ^ { - 2 x } \mathrm { dx } = \frac { 1 } { 2 } \mathrm { e } ^ { - 4 } < 0.0092∫2∞​e−2xdx=21​e−4<0.0092

Provide an Appropriate Response $\int _ { 2 } ^ { \infty } e ^ { - 2 x } \mathrm { dx } = \frac { 1 } { 2 } \mathrm { e } ^ { - 4 } < 0.0092$