Improper Integrals

**smartartbug** · May 20th 2010, 06:12 PM

Evaluate the integral or state that it diverges:

integral from 0 to 4 of dr/sqrt(4-r)

**AllanCuz** · May 20th 2010, 06:16 PM

Originally Posted by smartartbug

Evaluate the integral or state that it diverges:

integral from 0 to 4 of dr/sqrt(4-r)

$\int_0^4 \frac{1}{ \sqrt{4-r}}dr = -ln(4-r)_0^4 = -(ln(0) - ln(4))$

note that $ln(0)$ does not exist but

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

Therefore this integral diverges

**DrDank** · May 20th 2010, 08:37 PM

Originally Posted by AllanCuz

$\int_0^4 \frac{1}{ \sqrt{4-r}}dr = -ln(4-r)_0^4 = -(ln(0) - ln(4))$

note that $ln(0)$ does not exist but

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

Therefore this integral diverges

lolwut...?

$\int_0^4 \frac{1}{ \sqrt{4-r}}dr$

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

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

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

$\int_0^4u^{-\frac{1}{2}}du$

**AllanCuz** · May 20th 2010, 08:38 PM

Originally Posted by DrDank

lolwut...?

$\int_0^4 \frac{1}{ \sqrt{4-r}}dr$

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

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

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

$\int_0^4u^{-\frac{1}{2}}du$

oh **** I missed the square root...don't drink and integrate!

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