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

Question

Christopher Agresta · Accepted Answer

undefined To test the improper integral $\int_{-5}^{3} \frac{x}{|x|} \, dx$, we need to examine the behavior of the integrand as it approaches the endpoints of the interval $[-5, 3]$.

Since the function $\frac{x}{|x|}$ is not continuous at $x = 0$, we split the integral into two parts: $\int_{-5}^{0} \frac{x}{|x|} \, dx$ and $\int_{0}^{3} \frac{x}{|x|} \, dx$.

For the first part, $\int_{-5}^{0} \frac{x}{|x|} \, dx$, we have:
$$\int_{-5}^{0} \frac{x}{|x|} \, dx = \int_{-5}^{0} -1 \, dx = -\int_{-5}^{0} 1 \, dx = -(-5 - 0) = 5.$$

For the second part, $\int_{0}^{3} \frac{x}{|x|} \, dx$, we have:
$$\int_{0}^{3} \frac{x}{|x|} \, dx = \int_{0}^{3} 1 \, dx = 3 - 0 = 3.$$

Therefore, the integral $\int_{-5}^{3} \frac{x}{|x|} \, dx$ converges, and its value is $5 + 3 = 8$.

Aurora Alvarado · Answer

The integral does exist and equals #-2#.

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.