What makes an improper integral




















Otherwise, we say the improper integral diverges , which we capture in the following definition. First we compute the indefinite integral. Definition 2. Otherwise, we say the improper integral diverges. Evaluate it if it is convergent. With Example 2. For good measure, here is one more example.

We use the Comparison Test to show that it converges. Connect and share knowledge within a single location that is structured and easy to search.

I do not understand improper integrals. For both I need to evaluate the integral or show that it's divergent. I have no idea what to do and need serious help for math homework since we never covered this in our Calc 1 class.

Improper integrals occur in primarily two ways: an bound that goes off to infinity or a bound where the function goes off to infinity infinitely wide vs. Which should scream limit! So we set it up as follows:.

The other possibility is when the function in the integral does funky things on our bounds such as going off to infinity or dividing by zero, etc. One example is this simple integral:.

Now, to visit your examples: neither of them have an infinite bound and so our first type of improper integral can be ruled out. And we would want to solve:. In the current state, it seems that the second integral has no such "funk" and therefore is not improper. So we, instead, consider:.

Your second integral is proper since the denominator is always positive and never 0 on the interval of integration. The second integral can be done with a partial fraction decomposition. Sign up to join this community. This is a problem that we can do. So, this is how we will deal with these kinds of integrals in general. On a side note, notice that the area under a curve on an infinite interval was not infinity as we might have suspected it to be.

In fact, it was a surprisingly small number. We will call these integrals convergent if the associated limit exists and is a finite number i. Note as well that this requires BOTH of the integrals to be convergent in order for this integral to also be convergent.

If either of the two integrals is divergent then so is this integral. We can actually extend this out to the following fact. How fast is fast enough? The integral is then,. So, the first integral is convergent. Note that this does NOT mean that the second integral will also be convergent. This integral is convergent and so since they are both convergent the integral we were actually asked to deal with is also convergent and its value is,.

In most examples in a Calculus II class that are worked over infinite intervals the limit either exists or is infinite. These are integrals that have discontinuous integrands.

The process here is basically the same with one subtle difference.



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