# Evaluating INTEGRAL(-1, 1): 1^x2

• Dissonance in E
That is, the "improper" part of the integral, $\int_{-1}^1 x^{-2}dx$ is actually 0. The Cauchy principal value is 0 while the regular integral, which is what you calculated, does not exist.
Dissonance in E

## Homework Statement

Evaluate:
INTEGRAL(-1, 1): 1^x2

power rule

## The Attempt at a Solution

1/x^2
x^-2
x^-1
(-1/x)

plugging in values:
(-1/1) - (-1/-1)
-1 - 1
-2

Now i KNOW that -2 is an incorrect answer, the solution is infinity according to maple, nut I am not 100% why. I would guess its something to do with the fact that the integral involves the value x = 0 which is undefined for f(x) = -1/x as it asymptotically approaches 0.

Could i get a nice concise, clarification. thx

That needs to be done as an improper integral since the original function is not defined at x=0. You can't just integrate over 0, you have to approach it as a limit. Take the integral of 1/x^2 from a to 1 and let a->0 (similarly on the negative side).

ok so could i do it like this.

INT (-1, 1) : 1/x^2
=
INT (-1, A)(Lim A--> 0, from the left): 1/x^2
+
INT(A , 1)(Lim A -->0, from the right): 1/x^2

(-1/-A) -(-1/-1)
as A approaches 0 from the left we get
INFINITY - 1 = INFINITY

(-1/1) - (-1/A)
as A approaches 0 from the right we get
-1+INFINITY = INFINITY

Now by sum of the 2 limits
inf + inf = inf

Last edited:
Dissonance in E said:
ok so could i do it like this.

INT (-1, 1) : 1/x^2
=
INT (-1, A)(Lim A--> 0, from the left): 1/x^2
+
INT(A , 1)(Lim A -->0, from the right): 1/x^2

(-1/-A) -(-1/-1)
as A approaches 0 from the left we get
INFINITY - 1 = INFINITY

(-1/1) - (-1/A)
as A approaches 0 from the right we get
-1+INFINITY = INFINITY

Now by sum of the 2 limits
inf + inf = inf

Yes. That's the idea. Don't be too casual about doing arithmetic with 'infinity' though. E.g. infinity-infinity is not necessarily 0.

The "Cauchy principal value" for such an integral of a function not defined at 0 is
$$\lim_{\epsilon\to 0}\left[\int_{-1}^{-\epsilon} f(x) dx+ \int_{\epsilon}^1 f(x)dx[/itex] The difference between that and the "regular" integral is that we do not take the limits independently. With f(x)= 1/x2, we have [tex]\lim_{\epsilon\to 0}\left[-x^{-1}\right]_{-1}^{-\epsilon}+ \left[-x^{-1}\right]_\epsilon^1$$
$$= \lim_{\epsilon\to 0} \left(\frac{1}{\epsilon}- 1\right)+ \left(1- \frac{1}{\epsilon}\right)= 0$$.

## 1. What is the purpose of evaluating the integral (-1, 1): 1^x2?

The purpose of evaluating this integral is to find the area under the curve of the function 1^x2 between the limits of -1 and 1. This integral can also represent the antiderivative of the function 1^x2.

## 2. How is the integral (-1, 1): 1^x2 evaluated?

The integral (-1, 1): 1^x2 can be evaluated using integration techniques such as substitution, integration by parts, or the fundamental theorem of calculus. It can also be solved using numerical methods such as the trapezoidal rule or Simpson's rule.

## 3. What is the result of evaluating the integral (-1, 1): 1^x2?

The result of evaluating this integral is 2/3. This means that the area under the curve of the function 1^x2 between the limits of -1 and 1 is equal to 2/3, or the antiderivative of the function is 2/3x + C.

## 4. What does the value of the integral (-1, 1): 1^x2 represent?

The value of this integral represents the signed area under the curve of the function 1^x2 between the limits of -1 and 1. The signed area takes into account the positive and negative values of the function in the given interval.

## 5. How is the integral (-1, 1): 1^x2 related to the function 1^x2?

The integral (-1, 1): 1^x2 is the antiderivative of the function 1^x2. This means that if we differentiate the antiderivative, we will get the original function. In other words, the integral and the function are inverse operations of each other.

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