# Converge or Diverge?

• Sep 25th 2007, 09:06 PM
shilz222
Converge or Diverge?
Does the integral $\displaystyle \int_{1}^{\infty} \frac{3 \; dx}{2x \sqrt{4+5x}}$ converge or diverge?

If it converges, what is its value?
• Sep 25th 2007, 09:17 PM
Jhevon
Quote:

Originally Posted by shilz222
Does the integral $\displaystyle \int_{1}^{\infty} \frac{3x \; dx}{2x \sqrt{4+5x}}$ converge or diverge?

If it converges, what is its value?

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

We proceed by substitution:

Let $\displaystyle u = 4 + 5x$

$\displaystyle \Rightarrow du = 5 ~dx$

$\displaystyle \Rightarrow \frac 15~du = dx$

So our integral becomes:

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

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

I leave the rest to you
• Sep 25th 2007, 09:20 PM
Jhevon
Quote:

Originally Posted by shilz222
Its $\displaystyle \frac{3}{2} \int \frac{dx}{x \sqrt{4+5x}}$. You left out the $\displaystyle x$ in the denominator.

what are you talking about? you have 3x in the top and 2x in the bottom. the x's cancel
• Sep 25th 2007, 09:21 PM
shilz222
It should be just a 3 in the numerator.