Question

Q. ∫_(-∞)^∞▒e^(〖-x〗^2 ) dx=?
a. π/2
b. π
c. √π
d. 2√π

Answer 1

The given definite integration is a special integration ,

$$erf\left( x \right) = \frac{1}{{\sqrt \pi }}\int_{ - x}^x {{e^{ - {t^2}}}dt} = \frac{2}{{\sqrt \pi }}\int_{ 0}^x {{e^{ - {t^2}}}dt}\\ \int_{ - x}^x {{e^{ - {t^2}}}dt} = \frac{{\sqrt \pi }}{2}\int_{ - x}^x {\frac{{2{e^{ - {t^2}}}}}{{\sqrt \pi }}dt}$$

Now, use this concept and our integration will be

$$\int_{ - \infty }^\infty {{e^{ - {x^2}}}} dx = \frac{{\sqrt \pi }}{2}\int_{ - \infty }^\infty {\frac{{2{e^{ - {x^2}}}}}{{\sqrt \pi }}dx} \\ = \frac{{\sqrt \pi }}{2} \cdot \left[ {erf(x)} \right]_{ - \infty }^\infty \\ = \frac{{\sqrt \pi }}{2} \cdot 2\\ = \sqrt \pi$$

Therefore, the option C is the right answer.