Solve the problem.
-Given that
$$\int _ { 0 } ^ { \infty } e ^ { - x ^ { 2 } } d x = \frac { \sqrt { \pi } } { 2 } ,$$
evaluate
$$\int _ { 0 } ^ { \infty } e ^ { - 5 x ^ { 2 } } d x$$
A) $\frac { 1 } { 2 } \sqrt { \frac { \pi } { 5 } }$
B) $\frac { 1 } { 2 } \sqrt { 5 \pi }$
C) $\sqrt { \frac { \pi } { 5 } }$
D) $\sqrt { 5 \pi }$

Question

Quizplus · Accepted Answer

To evaluate $\int _ { 0 } ^ { \infty } e ^ { - 5 x ^ { 2 } } d x$, we can use the given integral $\int _ { 0 } ^ { \infty } e ^ { - x ^ { 2 } } d x = \frac { \sqrt { \pi } } { 2 }$ as a reference. By substituting $u = \sqrt{5}x$, we get $du = \sqrt{5}dx$, which means $dx = \frac{du}{\sqrt{5}}$. Substituting into the integral, we get $\int _ { 0 } ^ { \infty } e ^ { - u^2 } \frac{du}{\sqrt{5}}$, which simplifies to $\frac{1}{\sqrt{5}} \cdot \frac{\sqrt{\pi}}{2}$ after using the given integral. Simplifying further gives $\frac{1}{2} \sqrt{\frac{\pi}{5}}$.

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

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