How do you test the improper integral #int 2x^-3dx# from #[-1,1]# and evaluate if possible?

Question

Isaiah Allen · Accepted Answer

The integral diverges

To calculate the improper integral, proceed as follows #int_(-1)^1 2x^-3dx# #=lim_(t->O^-)(int_-1^t 2x^-3dx)+lim_(t->O^+)(int_t^1 2x^-3dx)# The first part is #lim_(t->O^-)(int_-1^t2x^-3dx)# #=lim_(t->O^-) [-1/x^2]_1^t# #=lim_(t->O^-)(1-1/t^2)# #=+oo# The integral diverges The second part is #lim_(t->O^+)(int_t^1x^-3dx)# #=lim_(t->O^+) [-1/x^2]_t^1# #=lim_(t->O^+)(1/t^2-1)# #=+oo# The integral diverges

William Abdalla · Answer

undefined To test the improper integral $ \int_{-1}^{1} 2x^{-3} dx $, we first need to determine if it converges or diverges. This integral is improper because it has an infinite discontinuity at $ x = 0 $. To test for convergence, we'll split the integral into two separate integrals: one from $ -1 $ to $ 0 $ and the other from $ 0 $ to $ 1 $. Then, we'll evaluate each integral separately.

For the integral from $ -1 $ to $ 0 $, we have:

$$ \int_{-1}^{0} 2x^{-3} dx = \lim_{a \to 0^-} \int_{-1}^{a} 2x^{-3} dx $$

Similarly, for the integral from $ 0 $ to $ 1 $, we have:

$$ \int_{0}^{1} 2x^{-3} dx = \lim_{b \to 0^+} \int_{b}^{1} 2x^{-3} dx $$

Now, let's evaluate each integral:

For the integral from $ -1 $ to $ 0 $:
$$ \lim_{a \to 0^-} \int_{-1}^{a} 2x^{-3} dx = \lim_{a \to 0^-} \left[ -\frac{2}{2x^2} \right]_{-1}^{a} $$
$$ = \lim_{a \to 0^-} \left( -\frac{1}{a^2} + \frac{1}{2} \right) = +\infty $$

For the integral from $ 0 $ to $ 1 $:
$$ \lim_{b \to 0^+} \int_{b}^{1} 2x^{-3} dx = \lim_{b \to 0^+} \left[ -\frac{2}{2x^2} \right]_{b}^{1} $$
$$ = \lim_{b \to 0^+} \left( -\frac{1}{1} + \frac{1}{2b^2} \right) = 1 $$

Since the integral from $ 0 $ to $ 1 $ converges to a finite value while the integral from $ -1 $ to $ 0 $ diverges to positive infinity, the original improper integral also diverges.