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Convergence and Divergence

shamieh

Active member
Sep 13, 2013
539
Determine whether the integral is Divergent or Convergent


\(\displaystyle \int^0_{-\infty} \frac{1}{3 - 4x} dx\)

I did a u substitution and got

\(\displaystyle \lim_{a\to\infty} -\frac{1}{4}\sqrt{3} + \frac{1}{4}\sqrt{3 - 4a}\)

So is because the \(\displaystyle -\infty\) is under the square root is it going to be divergent?

I have \(\displaystyle \frac{1}{4}\sqrt{3 - 4\infty}\)
 
Last edited:

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Determine whether the integral is Divergent or Convergent


\(\displaystyle \int^0_{-\infty} \frac{1}{3 - 4x} dx\)

I did a u substitution and got

\(\displaystyle \lim_{a\to\infty} -\frac{1}{4}\sqrt{3} + \frac{1}{4}\sqrt{3 - 4a}\)

So is because the \(\displaystyle -\infty\) is under the square root is it going to be divergent?

I have \(\displaystyle \frac{1}{4}\sqrt{3 - 4\infty}\)
Yes it is divergent.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
It is divergent, but not for the reason you cite. I would first use the substitution:

\(\displaystyle u=3-4x\,\therefore\,du=-4\,dx\) and our integral becomes:

\(\displaystyle I=-\frac{1}{4}\int_{\infty}^3\frac{du}{u}\)

Applying the rule \(\displaystyle -\int_b^a f(x)\,dx=\int_a^b f(x)\,dx\) we obtain:

\(\displaystyle I=\frac{1}{4}\int_3^{\infty}\frac{du}{u}\)

Now, since this is an improper integral, we may write:

\(\displaystyle I=\frac{1}{4}\lim_{t\to\infty}\left[\int_3^t\frac{du}{u} \right]\)

Applying the FTOC, and a property of logarithms, we find:

\(\displaystyle I=\frac{1}{4}\lim_{t\to\infty}\left(\ln\left(\frac{t}{3} \right) \right)=\infty\)

Thus, the given integral is divergent.