Question

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{1 / 2}^{\infty} \frac{1}{\sqrt{2 x-1}} d x $$

Short answer

The improper integral \(\int_{1 / 2}^{\infty} \frac{1}{\sqrt{2 x-1}} dx\) diverges.

Step-by-step solution

Transform the Improper Integral into Limit

First, replace \(\infty\) with \(t\) and then transform the improper integral into a limit as \(t\) approaches \(\infty\): \[ \lim_{t \to \infty} \int_{1/2}^{t} \frac{1}{\sqrt{2x-1}} dx \]

Calculate the Integral

The antiderivative of \(1/\sqrt{2x-1}\) can be found using substitution, by setting \(u=2x-1\). Thus, \[ \int_{1/2}^{t} \frac{1}{\sqrt{2x-1}} dx = \sqrt{u} = [2\sqrt{2x-1}]_{1/2}^t = 2\sqrt{2t-1} - 2 \]

Apply the Limit and Determine Convergence/Divergence

Now, apply the limit we've set up in step 1: \[ \lim_{t \to \infty} (2\sqrt{2t-1} - 2) \] As \(t\) approaches \(\infty\), \(2\sqrt{2t-1}\) goes to \(\infty\) and the whole expression becomes \(\infty\), meaning that the integral diverges.