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

1 Answer
Dec 26, 2017

The integral does exist and equals -2.

Explanation:

The function f(x)=x/|x| equals x/x = 1 if x > 0 and equals x/(-x)=-1 if x < 0. It is undefined at 0. The "impropriety" for the integral int_{-5}^{3}x/|x|dx therefore occurs at 0.

Assuming the integrals exist (converge), we can write int_{-5}^{3}x/|x|dx=lim_{b->0-}int_{-5}^{b}x/|x|dx+lim_{a->0+}int_{a}^{3}x/|x|dx (the notation means b approaches 0 "from the left" and a approaches 0 "from the right").

Now

lim_{b->0-}int_{-5}^{b}x/|x|dx=lim_{b->0-}int_{-5}^{b}(-1)dx=lim_{b->0-}(-x)|_{-5}^{b}

=lim_{b->0-}(-b-(-(-5)))=lim_{b->0-}(-b-5)=-5

and

lim_{a->0+}int_{a}^{3}x/|x|dx=lim_{a->0+}int_{a}^{3}1dx=lim_{a->0+}(x)|_{a}^{3}

=lim_{a->0+}(3-a)=3.

Therefore, these integrals converge and

int_{-5}^{3}x/|x|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.