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How do you test the improper integral #int x^-2dx# from #[-1,1]# and evaluate if possible?

Answer 1

Charles Arocha

The improper integral:

#int_(-1)^1 x^(-2)dx#

is divergent.

Consider first that #x^-2# is an even function:

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

so that:

#int_(-1)^1 x^(-2)dx = 2int_0^1 x^-2dx#

Now pose:

#f(t) = int_t^1 x^-2dx = [x^-1/(-1)]_t^1 = -1+1/t= (1-t)/t#

So that:

#int_0^1 x^(-2)dx = lim_(t->0^+) f(t) = lim_(t->0^+) (1-t)/t = +oo#

then the improper integral:

#int_(-1)^1 x^(-2)dx#

is divergent.

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Answer 2

Paisley Johnson

To test the improper integral ∫x^(-2)dx from -1 to 1, first, identify if there are any singularities within the interval of integration. In this case, x^(-2) becomes undefined at x = 0, which is within the interval [-1, 1]. Then, split the integral into two parts: ∫x^(-2)dx from -1 to 0 and ∫x^(-2)dx from 0 to 1. Evaluate each part separately. The integral of x^(-2)dx is -x^(-1), so integrating from -1 to 0 yields -(-1)^(-1) - (-0)^(-1) = -(-1) - (-∞) = 1 + ∞ = ∞. Integrating from 0 to 1 yields (-1)^(-1) - (0)^(-1) = -1 - ∞ = -∞. Since both parts diverge, the original improper integral diverges.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.