Question #91367
Q. ∫_(-∞)^∞▒e^(〖-x〗^2 ) dx=?
a. π/2
b. π
c. √π
d. 2√π
1
Expert's answer
2019-07-09T11:51:49-0400

The given definite integration is a special integration ,

erf(x)=1πxxet2dt=2π0xet2dtxxet2dt=π2xx2et2πdterf\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

ex2dx=π22ex2πdx=π2[erf(x)]=π22=π\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.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS