Determine whether the integral from minus infinity to 0 of 1/(3 times the square root of (x - 8)^2) dx converges, and evaluate it if it does.

Step-by-step solution

1. 1

   Turn the square root into an absolute value. For every real $x$, $\sqrt{(x-8)^2}=|x-8|$ — not $x-8$. This distinction is the whole point of the problem, because the integration range sits far to the left of $8$.

2. 2

   Drop the absolute value on the actual interval. On $(-\infty,0]$ we have $x\le 0<8$, so $x-8<0$ and $|x-8|=8-x$. The integral becomes $\int_{-\infty}^{0}\frac{dx}{3(8-x)}.$

3. 3

   Write it as a limit. The lower endpoint is infinite, so by definition $\int_{-\infty}^{0}\frac{dx}{3(8-x)}=\lim_{a\to-\infty}\int_{a}^{0}\frac{dx}{3(8-x)}.$ The integrand itself is continuous and positive on the whole range — the only source of trouble is the unbounded interval.

4. 4

   Find an antiderivative. Substituting $u=8-x$, $du=-dx$, gives $\int\frac{dx}{3(8-x)}=-\frac13\int\frac{du}{u}=-\frac13\ln|8-x|+C.$

5. 5

   Evaluate the limit. $\lim_{a\to-\infty}\left[-\frac13\ln(8-x)\right]_{a}^{0}=\lim_{a\to-\infty}\left(-\frac13\ln 8+\frac13\ln(8-a)\right).$ As $a\to-\infty$, $8-a\to+\infty$ and $\ln(8-a)\to+\infty$, so the expression grows without bound.

6. 6

   Conclude. The integral diverges to $+\infty$. This is the usual $1/x$ tail: the integrand decays like $\frac{1}{3|x|}$, and $\int^{\infty}\frac{dx}{x}$ is the borderline case that fails to converge.