Improper Integrals

Dec 2009
11
0
Evaluate the integral or state that it diverges:

integral from 0 to 4 of dr/sqrt(4-r)
 
Apr 2010
384
153
Canada
Evaluate the integral or state that it diverges:

integral from 0 to 4 of dr/sqrt(4-r)
\(\displaystyle \int_0^4 \frac{1}{ \sqrt{4-r}}dr = -ln(4-r)_0^4 = -(ln(0) - ln(4)) \)

note that \(\displaystyle ln(0) \) does not exist but

\(\displaystyle \lim_{x \to 0 } ln(x) = - \infty \)

Therefore this integral diverges
 
  • Like
Reactions: smartartbug
May 2010
20
8
\(\displaystyle \int_0^4 \frac{1}{ \sqrt{4-r}}dr = -ln(4-r)_0^4 = -(ln(0) - ln(4)) \)

note that \(\displaystyle ln(0) \) does not exist but

\(\displaystyle \lim_{x \to 0 } ln(x) = - \infty \)

Therefore this integral diverges
lolwut...?

\(\displaystyle \int_0^4 \frac{1}{ \sqrt{4-r}}dr\)

let \(\displaystyle u=4-r\) therefore \(\displaystyle du=(-1)dx\)

\(\displaystyle \int_4^0 \frac{1}{ \sqrt{u}}(-1)du\)

\(\displaystyle -\int_4^0 \frac{1}{ \sqrt{u}}du\)

\(\displaystyle \int_0^4u^{-\frac{1}{2}}du\)
 
  • Like
Reactions: AllanCuz
Apr 2010
384
153
Canada
lolwut...?

\(\displaystyle \int_0^4 \frac{1}{ \sqrt{4-r}}dr\)

let \(\displaystyle u=4-r\) therefore \(\displaystyle du=(-1)dx\)

\(\displaystyle \int_4^0 \frac{1}{ \sqrt{u}}(-1)du\)

\(\displaystyle -\int_4^0 \frac{1}{ \sqrt{u}}du\)

\(\displaystyle \int_0^4u^{-\frac{1}{2}}du\)
oh **** I missed the square root...don't drink and integrate!
 
Similar Math Discussions Math Forum Date
Calculus
Calculus
Calculus
Calculus