Convergent/Divergent

K

klooless

New member
Joined
Jun 10, 2009
Messages
19
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!
 
Be careful. \(\displaystyle du=\frac{1}{2\sqrt{x}}dx\)
 
klooless said:
antiderivative of 1/sqr(x) * e^-sqr(x) from [0, 3]
Click to expand...
Here is a suggestion. What is the derivative of \(\displaystyle - 2e^{ - \sqrt x }\)?
 
klooless said:
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!
Click to expand...

I get

\(\displaystyle \int^3_0\frac{1}{\sqrt{x}}e^{-\sqrt{x}}dx \, = \, 2\int^{\sqrt{3}}_0e^{-u} du \,\)
 
\(\displaystyle Is \ that \ a \ = \ \int_{0}^{3}\frac{dx}{\sqrt(x) e^{-\sqrt(x)}} \ or \ b \ = \ \int_{0}^{3}\frac{e^{-\sqrt(x)}dx}{\sqrt(x)}\)

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

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

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).
 
\(\displaystyle \left. { - 2e^{ - \sqrt x } } \right|_0^3 = - 2e^{ - \sqrt 3 } + 2\)
 
You must log in or register to reply here.
Top