# How do you test the improper integral int x/absxdx from [-5,3] and evaluate if possible?

Dec 26, 2017

The integral does exist and equals $- 2$.

#### Explanation:

The function $f \left(x\right) = \frac{x}{|} x |$ equals $\frac{x}{x} = 1$ if $x > 0$ and equals $\frac{x}{- x} = - 1$ if $x < 0$. It is undefined at 0. The "impropriety" for the integral ${\int}_{- 5}^{3} \frac{x}{|} x | \mathrm{dx}$ therefore occurs at $0$.

Assuming the integrals exist (converge), we can write ${\int}_{- 5}^{3} \frac{x}{|} x | \mathrm{dx} = {\lim}_{b \to 0 -} {\int}_{- 5}^{b} \frac{x}{|} x | \mathrm{dx} + {\lim}_{a \to 0 +} {\int}_{a}^{3} \frac{x}{|} x | \mathrm{dx}$ (the notation means $b$ approaches 0 "from the left" and $a$ approaches 0 "from the right").

Now

${\lim}_{b \to 0 -} {\int}_{- 5}^{b} \frac{x}{|} x | \mathrm{dx} = {\lim}_{b \to 0 -} {\int}_{- 5}^{b} \left(- 1\right) \mathrm{dx} = {\lim}_{b \to 0 -} \left(- x\right) {|}_{- 5}^{b}$

$= {\lim}_{b \to 0 -} \left(- b - \left(- \left(- 5\right)\right)\right) = {\lim}_{b \to 0 -} \left(- b - 5\right) = - 5$

and

${\lim}_{a \to 0 +} {\int}_{a}^{3} \frac{x}{|} x | \mathrm{dx} = {\lim}_{a \to 0 +} {\int}_{a}^{3} 1 \mathrm{dx} = {\lim}_{a \to 0 +} \left(x\right) {|}_{a}^{3}$

$= {\lim}_{a \to 0 +} \left(3 - a\right) = 3$.

Therefore, these integrals converge and

${\int}_{- 5}^{3} \frac{x}{|} x | \mathrm{dx} = - 5 + 3 = - 2$.

In the end, you should know that a finite number of "jump discontinuities " over a finite interval will not cause an improper integral over that interval to diverge.