Integral of x^-2 b/w -1 & 1: -2 Explained

In summary, the integral of 1/x^2 between -1 and 1 does not equal -2 because the function is not differentiable in that range due to an infinite discontinuity at 0. The improper integrals approach can be used to evaluate the integral, but the limit does not exist as the limits of integration approach 0 independently. If the integral exists, the Cauchy Principle Value can be used, but in this case it does not exist and results in a nonsensical answer.
  • #1
Zoe-b
98
0

Homework Statement


The integral of 1/(x^2) between -1 and 1 comes to -2. How is this possible when the graph is always above the x-axis?

3. The Attempt at a Solution
the integral of x^(-2) = -x^(-1) + c
in this range you get (-1/1 - 1/1) which does indeed equal 2.
My original solution was to say that since the graph is not continuous in this range it makes no sense to try to find the integral (which should be infinite surely?). Is there more to it than this?
Thanks
 
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  • #2
Short answer: the integral of [itex]1/x^2[/itex] between -1 and 1 does NOT "come to -2". In fact, [itex]1/x^2[/itex] is not differentiable between -1 and 1 because of the infinite discontinuity at 0. Specifically, you cannot just find the anti-derivative and evaluate at -1 and 1.

You could try to evaluate it using the definition of "improper integrals":
[tex]\int_{-1}^1 \frac{1}{x^2} dx= \int_{-1}^{-\alpha}\frac{1}{x^2}dx+ \int_\beta^1 \frac{1}{x^2}dx[/tex]
and then take the limits as [itex]\alpha[/itex] and [itex]\beta[/itex] go to 0independently.
The two integrals are
[tex]-\frac{1}{x}\left|_{-1}^{-\alpha}+ \frac{1}{x}\right|_{\beta}^1= \frac{1}{\alpha}- 1- 1+ \frac{1}{\beta}[/tex]
but that limit does not exist as [itex]\alpha[/itex] and [itex]\beta[/itex] go to 0 independently
.
If you were to assume that [itex]\alpha= \beta[/itex] you would be calcuating the "Cauchy Principle Value"- those two fractions would cancel and you would get "-2". If the integral itself exists, then the Cauchy Principle Value but if the integral does not exist, as here, you get the kind of ridiculous result you cite.
 
  • #3
Ok that's fine, pretty much what I thought (though hadn't thought through the limits thing)
Thank you
 
  • #4
HallsofIvy said:
Short answer: the integral of [itex]1/x^2[/itex] between -1 and 1 does NOT "come to -2". In fact, [itex]1/x^2[/itex] is not differentiable between -1 and 1 because of the infinite discontinuity at 0. Specifically, you cannot just find the anti-derivative and evaluate at -1 and 1.

You could try to evaluate it using the definition of "improper integrals":
[tex]\int_{-1}^1 \frac{1}{x^2} dx= \int_{-1}^{-\alpha}\frac{1}{x^2}dx+ \int_\beta^1 \frac{1}{x^2}dx[/tex]
and then take the limits as [itex]\alpha[/itex] and [itex]\beta[/itex] go to 0independently.
The two integrals are
[tex]-\frac{1}{x}\left|_{-1}^{-\alpha}+ \frac{1}{x}\right|_{\beta}^1= \frac{1}{\alpha}- 1- 1+ \frac{1}{\beta}[/tex]
but that limit does not exist as [itex]\alpha[/itex] and [itex]\beta[/itex] go to 0 independently
.

If you were to assume that [itex]\alpha= \beta[/itex] you would be calcuating the "Cauchy Principle Value"- those two fractions would cancel and you would get "-2".
. Actually, they wouldn't cancel; they both go to [itex]\infty[/itex]. If the integrand was odd they would have cancelled.
If the integral itself exists, then the Cauchy Principle Value but if the integral does not exist, as here, you get the kind of ridiculous result you cite.
 

Related to Integral of x^-2 b/w -1 & 1: -2 Explained

1. What is the integral of x^-2 between -1 and 1?

The integral of x^-2 between -1 and 1 is equal to -2. This is calculated by taking the antiderivative of x^-2, which is x^-1, and then evaluating it at the upper and lower bounds, -1 and 1. This results in (-1)^-1 - (1)^-1 = -2.

2. How is the integral of x^-2 between -1 and 1 equal to -2?

The integral of x^-2 between -1 and 1 is equal to -2 because the antiderivative of x^-2 is x^-1, and when evaluated at the upper and lower bounds, -1 and 1, it results in (-1)^-1 - (1)^-1 = -2.

3. Why is the integral of x^-2 between -1 and 1 negative?

The integral of x^-2 between -1 and 1 is negative because the function x^-2 is negative between -1 and 1, meaning it is below the x-axis. When calculating the definite integral, the area under the curve is taken into consideration, and since the function is below the x-axis, the area is negative.

4. What is the significance of the integral of x^-2 between -1 and 1?

The integral of x^-2 between -1 and 1 has a significance in calculus as it represents the total area under the curve of the function x^-2 between the given bounds. It is also used to calculate important values such as average value and volume of revolution in certain applications.

5. Can the integral of x^-2 between -1 and 1 be approximated?

Yes, the integral of x^-2 between -1 and 1 can be approximated using numerical methods such as the trapezoidal rule or Simpson's rule. These methods divide the area under the curve into smaller trapezoids or parabolas and sum them up to approximate the total area. The more subdivisions used, the more accurate the approximation will be.

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