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My thinking started when I got the answer to below integral $$\int_{-2}^{1}\frac{1}{x^2}dx$$

I normally found the anti-derivative as $\frac{-1}{x}$ and just substituted the limit. The answer I was getting is $\frac{-3}{2}$. Suddenly I thought "why would I get a negative value for this integral. The integrand is positive for all real values, then why would I get a negative answer?"

The question for which I was solving this integral asked to also plot the area for of this integral. Thus, I have also looked at the graph then I remembered that this function is discontinuous at $x=0$ which is inside the interval of the limits.

Now, my question is how did I even get the answer $-3/2$ and how can I evaluate this integral?

I honestly think the question is incorrect and the limits should be different. Cus it approaches infinity at $x=0$ thus area won't be a finite value. Am I correct?

ACB
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Rayon Cherry
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  • Approaching infinity at $x=0$ is an issue, but it may matter how you approach infinity and the method of integration you are using. $\int\limits_{-2}^{1}\frac{1}{\sqrt{|x|}},dx$ does have a plausible finite value of $2(\sqrt{2}+1) \approx 4.8284$ but your example does not – Henry Feb 23 '22 at 11:58

2 Answers2

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The area is infinite. The point of the problem is to demonstrate to you that integration is defined (so far) only for bounded functions on bounded intervals (and the functions need to be piece-wise continuous.) Since your integrand isn't bounded, the result you got $-3/2$ is meaningless.

This problem is probably your book's or instructor's introduction to the topic of improper integrals.

B. Goddard
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@B.Goddard has provided the correct response to your question. I would like to point of that there is a sense in which the value of this integral is $-3/2$.

Consider $$ f(t)=\int_{-2}^1\frac{\mathrm dx}{x-t}. $$ so that in some sense we may write $$ f^\prime(0)=\int_{-2}^1\frac{\mathrm dx}{x^2}, $$ which is your integral. Integrating we find $$ f(t)=\ln(1-t)-\ln(-2-t), $$ which upon differentiating and passing to the limit gives $$ f^\prime(0)=-\frac{3}{2}\implies \int_{-2}^1\frac{\mathrm dx}{x^2}``="-\frac{3}{2}. $$ I have placed quotes here because this is not a true equality. Indeed, there are ways to assign values to divergent integrals in a similar way as presented here. For more information, take a look at the Cauchy principal value, and Hadamard regularization.

Aaron Hendrickson
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