Improper integral

  • Thread starter ggcheck
  • Start date
  • #1
87
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I used u substitution to get to this point:

[tex]\lim_{R\rightarrow\ 3-}} \int_{-3}^{R-3} (\frac{u + 3}{sqrt{u}}) du[/tex]

is the only way to proceed from here using integration by parts?
 
Last edited:

Answers and Replies

  • #2
1,631
4
I used u substitution to get to this point:

[tex]\lim_{R\rightarrow\ 3-}} \int_{-3}^{R-3} (\frac{u + 3}{sqrt{u}}) du[/tex]

is the only way to proceed from here using integration by parts?

If your calculations are correct up to this point, i do not think you need to use integ by parts. Simply just try to rearrange the integrand like this

(u+3)/u^1/2= u/u^1/2 +3/u^1/2 = u^1-1/2 +3u^-1/2 and i guess you will be fine!!!!
 
  • #3
87
0
If your calculations are correct up to this point, i do not think you need to use integ by parts. Simply just try to rearrange the integrand like this

(u+3)/u^1/2= u/u^1/2 +3/u^1/2 = u^1-1/2 +3u^-1/2 and i guess you will be fine!!!!
ahhh, I didn't think to break up the fraction, for some reason I missed that

thanks :)
 

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