Convergent/Divergent

[klooless]

antiderivative of 1/sqr(x) * e^-sqr(x) from [0, 3]

first I substituted u= sqr(x) and du= 2x dx ect.

Then I used uv- integ( du v ) with u= e^-u, and dv= u^-3 du. My result was more complicated then I expected, is this the right way to figure out if it's convergent/divergent?

A push in the right direction would be great. Cheers!

 

[galactus]

Be careful. $\displaystyle du=\frac{1}{2\sqrt{x}}dx$

 

[pka]

antiderivative of 1/sqr(x) * e^-sqr(x) from [0, 3]

Here is a suggestion. What is the derivative of $\displaystyle - 2e^{ - \sqrt x }$?

 

[Deleted member 4993]

antiderivative of 1/sqr(x) * e^-sqr(x) from [0, 3]

first I substituted u= sqr(x) and du= 2x dx ect. <<< I am sure you meant dx = 2u du

Then I used uv- integ( du v ) with u= e^-u, and dv= u^-3 du. My result was more complicated then I expected, is this the right way to figure out if it's convergent/divergent? <<< Does not look correct

A push in the right direction would be great. Cheers!

I get

$\displaystyle \int^3_0\frac{1}{\sqrt{x}}e^{-\sqrt{x}}dx \, = \, 2\int^{\sqrt{3}}_0e^{-u} du \,$

 

[BigGlenntheHeavy]

$\displaystyle Is \ that \ a \ = \ \int_{0}^{3}\frac{dx}{\sqrt(x) e^{-\sqrt(x)}} \ or \ b \ = \ \int_{0}^{3}\frac{e^{-\sqrt(x)}dx}{\sqrt(x)}$

$\displaystyle In \ either \ case: \ a \ = \ 2\int_{0}^{\sqrt3}\frac{du}{e^{-u}} \ or \ b \ = \ 2\int_{0}^{\sqrt3}e^{-u}du$

$\displaystyle In \ both \ cases \ let \ u \ = \ x^{1/2}, \ then \ u^{2} \ = \ x \ and \ 2udu \ = \ dx.$

Note: I just noticed that in both cases, you are dealing with an improper integral, however, not to worry as both converge (e^0=1).

 

[klooless]

Thanks all!

I got -2e^-sqr(3) therefore convergent, yes?

(Glenn it was b)

 

[pka]

$\displaystyle \left. { - 2e^{ - \sqrt x } } \right|_0^3 = - 2e^{ - \sqrt 3 } + 2$

 

[klooless]

Excellent, thanks!