how do i find the anti derivative of the definate integral from (-1,1) of 1/(x^2)?

Question

how do i find the anti derivative of the definate integral from (-1,1) of 1/(x^2).

Answer 1

Note: This is an improper integral in disguise. How you know this is because x is not defined at 0, and your interval goes from -1 to 1 (which includes 0!) For that reason, you have to take two separate integrals: -1 to 0-, and 0+ to 1. When I write 0- and 0+, I mean "0 from the left" and "0 from the right", because this will involve taking limits.

A = lim (t -> 0-, Integral (-1 to t-, 1/x^2)dx) + lim (t -> 0+, Integral (t+ to 1, 1/x^2)dx)

Note that the integral of 1/x^2 is equal to -1/x. We evaluate it from -1 to t, and from t to 1, like so:

A = lim (t -> 0-, [-1/t - -1/-1]) + lim (t -> 0+, -1/1 - -1/t)
A = lim (t -> 0-, [-1/t - 1]) + lim (t -> 0+, -1 + 1/t)

A = infinity + infinity = infinity.

Answer 2

1/(x^2) can be rewritten as x^-2 which makes it easier.
The antierivative is -x^-1 or -1/x.

Since the antiderivative has an x in the denominator, and x equals 0 can not be evaluated, because 1/0 is undefined. It makes this an improper integral. Therefoer this integral or anti derivative has to be handled by taking the limit as x approaches infinity.

Hope this helps.

Answer 3

once you're taking the spinoff of a fundamental all you do is plug the better decrease (x thus) in to the equation the position t is. Then distinct the hot time period with techniques from the spinoff of x. So your answer is: (sinx^2)(a million) yet on the grounds that your multiplying with techniques from one you could go away that off. Your answer is sin(x^2)

Answer 4

so, you want the integral of the integral? errr???
integral of 1/x^2 dx = integral of x^-2 dx = -1/x +C ....... if you integrate from -1 to 1, you get -1 - (1) = -2
if you want the anti-derivative of that, well, integral of -2 dx = -2x +C