Solve the problem.
-Given that
$$\int _ { 0 } ^ { \infty } \mathrm { e } ^ { - \mathrm { x } ^ { 2 } } \mathrm { dx } = \frac { \sqrt { \pi } } { 2 } ,$$
evaluate
$$\int _ { - \infty } ^ { \infty } e ^ { - 6 x ^ { 2 } } d x$$
A) $\frac { 1 } { 2 } \sqrt { 6 \pi }$
B) $\sqrt { \frac { \pi } { 6 } }$
C) $\sqrt { 6 \pi }$
D) $\frac { 1 } { 2 } \sqrt { \frac { \pi } { 6 } }$

Question

Quizplus · Accepted Answer

To evaluate $\int _ { - \infty } ^ { \infty } e ^ { - 6 x ^ { 2 } } d x$, we can use a substitution to relate it to the given integral. Let $u = \sqrt{6}x$, then $du = \sqrt{6}dx$, and $dx = \frac{du}{\sqrt{6}}$. Substituting into the integral, we get $\int _ { - \infty } ^ { \infty } e ^ { - u^2 } \frac{du}{\sqrt{6}}$, which equals $\frac{1}{\sqrt{6}}\int _ { - \infty } ^ { \infty } e ^ { - u^2 } du$. Given $\int _ { 0 } ^ { \infty } e ^ { - x ^ { 2 } } dx = \frac { \sqrt { \pi } } { 2 }$, the integral from $-\infty$ to $\infty$ is twice this, or $\sqrt{\pi}$. Thus, the answer is $\sqrt{\pi}/\sqrt{6} = \sqrt{\frac{\pi}{6}}$.

Solve the Problem ∫0∞e−x2dx=π2,\int _ { 0 } ^ { \infty } \mathrm { e } ^ { - \mathrm { x } ^ { 2 } } \mathrm { dx } = \frac { \sqrt { \pi } } { 2 } ,∫0∞​e−x2dx=2π​​,

Solve the Problem $\int _ { 0 } ^ { \infty } \mathrm { e } ^ { - \mathrm { x } ^ { 2 } } \mathrm { dx } = \frac { \sqrt { \pi } } { 2 } ,$