# Improper Integrals

• May 20th 2010, 06:12 PM
smartartbug
Improper Integrals
Evaluate the integral or state that it diverges:

integral from 0 to 4 of dr/sqrt(4-r)
• May 20th 2010, 06:16 PM
AllanCuz
Quote:

Originally Posted by smartartbug
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
• May 20th 2010, 08:37 PM
DrDank
Quote:

Originally Posted by AllanCuz
$\displaystyle \int_0^4 \frac{1}{ \sqrt{4-r}}dr = -ln(4-r)_0^4 = -(ln(0) - ln(4))$

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 $\displaystyle du=(-1)dx$

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

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

$\displaystyle \int_0^4u^{-\frac{1}{2}}du$
• May 20th 2010, 08:38 PM
AllanCuz
Quote:

Originally Posted by DrDank
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!