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$$\int_{-\infty}^{2} x^{-4}dx$$

Since its type 1 I let A = $-\infty$

$$\lim_{A\to-\infty} \int_{A}^{2} x^{-4}dx $$

$$\lim_{A\to-\infty} \frac{x^{-3}}{-3}\bigg|_{A}^{2} = \lim_{A\to-\infty} -1/3\big(\frac{1}{8} - \frac{1}{A^3}\big) = -\frac{1}{24} - 0 = -1/24$$

Why does wolfram tell me this is undefined?

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    The function has a singularity at $x=0$. – 2017-02-16
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    OHHHHH. I completely forgot. Thank you – 2017-02-16
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    Plus, although this is not the case this time, I wouldn't accept the answers from wolfram alpha as correct.I have seen lots of wrong answers from it. – 2017-02-16

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The integral $\int_{-\infty}^{2} x^{-4}dx$ is divergent, since $\int_{0}^{2} x^{-4}dx$ diverges.