# Improper Integrals

#### smartartbug

Evaluate the integral or state that it diverges:

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

#### AllanCuz

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

smartartbug

#### DrDank

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

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 $$\displaystyle du=(-1)dx$$

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

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

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

AllanCuz

#### AllanCuz

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!

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