Help me with this integral problem?

Question

Show whether ∫ 1/x^4 dx from -1 to 2 is convergent or divergent. If convergent, evaluate the integral.

Here's my work:

After integrating, I got [-1/(3x^3)] from -1 to 2 and I got -3/8 as the answer for the integral. But I don't know why the answer says it's divergent. Can anyone please tell me why this integral is divergent and correct me? Also, how can I show if it's divergent?

Answer 1

Hi Sun,

So since the question is asking about whether or not a definite integral is convergent or not, that tells me that we might be dealing with an improper integral.

For our purposes, there are usually two types of improper integrals:

1) Either one or both of the limits of integration are ∞ or -∞

2) The function is discontinuous on the interval that it is being integrated over.

It does not seem to be case 1, since the limits are from -1 to 2, so can we check if its case 2?

Is f(x) = 1/x⁴ continuous over the interval [-1,2]? Maybe sketching a graph and seeing how that looks can help you.

I hope this helps you get to the answer. I'll check back if you have any follow up questions.

Answer 2

integral [ 1/x^4] = integral [ x^(-4)] = (-1/3)x^(-3)

because the asymptote at zero, it must be integrated in 2 parts:

from -1 to 0 and 0 to 2...

BUT

lim [ (-1/3)x^(-3) ] as x-->0 does not exist

Answer 3

Looking at the graph shows you why.

Trying to find the area under the curve when the curve grows unbounded between -1 ad 0 and then again between 0 and 2 is not possible