Evaluate the improper intergral:

Question

anonymous · Answer

$$\int\limits_{0}^{1}x ^{\frac{ -1 }{ 2 }}dx$$

reemii · Answer

the method is to compute first $\int_\varepsilon^1 x^{-1/2} \,dx$. 
This will be an expression involving the $\varepsilon$. then compute the limit for of this function of $\varepsilon$ when $\varepsilon\to0+$.

anonymous · Answer

$$\int\limits_{-\infty}^{\infty} xe ^{-x ^{2}}dx $$  @reemii !!! Evaluate the improper integral

reemii · Answer

isn't the definition: this integral exists if both improper integrals exist ?
$\int_{-\infty}^0 f(x)dx$ and $\int_0^{\infty} f(x)dx$

reemii · Answer

yes it's that.
So, concentrating on $\int_0^\infty x e^{-x^2}dx$, .. where do you think the $x$ is coming from?

anonymous · Answer

I have no idea what ur asking

reemii · Answer

oh, I mean, if you compute the derivative of $e^{-x^2}$ , you'll obtain $e^{-x^2}\times x\times C$. So, .. integrating this function is very easy.

reemii · Answer

C=-2

anonymous · Answer

I have an answere sheet and it says the answere is 0....

reemii · Answer

yes it's correct. the integral on the right is $[-\frac12 e^{-x^2}]_0^\infty = 0  -   (-1)$. On the "left", it's equal to -1. therefore the complete integral is -1+1=  0.

anonymous · Answer

Ohhh. ok!

reemii · Answer

oops, rather:  -1/2 + 1/2 = 0 ^^'