0
$\begingroup$

$$\int_{-2}^{8}\dfrac{dx}{\sqrt{|2x\|}}$$

I understand that you have to split this into two integrals because at $x=0$, the function is not defined. The example showed that they split up the integral like this:

$$\lim_{b\to0^{-}}\int_{-2}^{b}\dfrac{dx}{\sqrt{-2x}}+\lim_{c\to0^{+}}\int_{c}^{8}\dfrac{dx}{\sqrt{2x}}$$

I understand how they split up the integral but why is the denominator different in each one? I figured that the first integral has negative limits of integration so a negative has to be put in the square root to make it positive. The integral on the right hand side remains positive because the limits are never negative. Is this the right assumption?

If my assumption is correct, is there a general rule for dealing with absolute values in problems like these? I appreciate anyone that would take the time to explain this to me. Thank you in advance!

$\endgroup$
  • 1
    $\begingroup$ When $t$ is negative, $|t|=-t$. This simple "trick" (really, part of the definition of absolute value) is often useful. $\endgroup$ – André Nicolas Oct 21 '13 at 3:18
1
$\begingroup$

If the integral involves $|x|$, we split the interval of integration by $0$, and unwind the definition of $|x|$ on each part: $|x|=x$ on the positive part, $|x|=-x$ on the negative. Then integrate and add the results. If only the convergence of an improper integral is of interest, we don't calculate the values, but only check whether each part converges.

More generally: if the integral involves $|g(x)|$ where $g$ is some function of $x$, the interval of integration is split by the points where $g(x)$ changes sign.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.