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

The integral diverges

To calculate the improper integral, proceed as follows

The first part is

The integral diverges

The second part is

The integral diverges

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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.

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