Converge or Diverge?

**shilz222** · September 25th 2007, 09:06 PM

Does the integral $\int_{1}^{\infty} \frac{3 \; dx}{2x \sqrt{4+5x}}$ converge or diverge?

If it converges, what is its value?

**Jhevon** · September 25th 2007, 09:17 PM

Originally Posted by shilz222

Does the integral $\int_{1}^{\infty} \frac{3x \; dx}{2x \sqrt{4+5x}}$ converge or diverge?

If it converges, what is its value?

$\int_{1}^{\infty} \frac{3x}{2x \sqrt{4+5x}}~dx = \frac 32 \lim_{N \to \infty} \int_{1}^{N} \frac 1{\sqrt {4 + 5x}}~dx$

We proceed by substitution:

Let $u = 4 + 5x$

$\Rightarrow du = 5 ~dx$

$\Rightarrow \frac 15~du = dx$

So our integral becomes:

$\frac 3{10} \lim_{N \to \infty} \int_{x = 1}^{x = N} u^{- \frac 12}~du = \frac 3{10} \left[ 2 u^{\frac 12} \right]_{x = 1}^{x = N}$

$= \frac 3{10} \lim_{N \to \infty} \left[ 2 \sqrt {4 + 5x} \right]_{1}^{N}$

I leave the rest to you

**Jhevon** · September 25th 2007, 09:20 PM

Originally Posted by shilz222

Its $\frac{3}{2} \int \frac{dx}{x \sqrt{4+5x}}$ . You left out the $x$ in the denominator.

what are you talking about? you have 3x in the top and 2x in the bottom. the x's cancel

**shilz222** · September 25th 2007, 09:21 PM

It should be just a 3 in the numerator.

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