### Video Transcript

Determine whether the integral from negative β to β of π¦ cubed minus three π¦ squared with respect to π¦ is convergent or divergent.

In this question, weβre given a definite integral and weβre asked to determine whether this integral is convergent or divergent. The first thing we need to notice about our integral is itβs improper. We say that an integral is improper if either its upper or lower limit is equal to positive or negative β. And in this case, both the upper and lower limit of integration are a form of β. So we need to recall how we evaluate integrals in this form. So weβll start by recalling our rules for integrating improper integrals in this form.

We recall the integral from negative β to β of π of π¦ with respect to π¦ will be equal to the limit as π approaches negative β of the integral from π to π of π of π¦ with respect to π¦ plus the limit as π‘ approaches β of the integral from π to π‘ of π of π¦ with respect to π¦. And thereβs a few things worth pointing out about this definition. First, we can choose any real constant for our value of π, although usually the easiest choice is to choose the constant zero. Next, this definition is only valid if both of our limits converge. If either of them diverge, then we say our original integral is divergent. And this is what the question means when it asks us about the convergence or divergence of this integral. We need to determine the convergence or divergence of these limits.

So letβs try applying this to the integral given to us in the question. Weβll choose our value of π equal to zero. Doing this, we get the improper integral given to us in the question is equal to the limit as π approaches negative β of the integral from π to zero of π¦ cubed minus three π¦ squared with respect to π¦ plus the limit as π‘ approaches β of the integral from zero to π‘ of π¦ cubed minus three π¦ squared with respect to π¦, provided both of these limits exist. And we can evaluate both of these integrals by using the power rule for integration. We want to add one to our exponent of π¦ and then divide by this new exponent.

So weβll start by integrating π¦ cubed with respect to π¦. We add one to the exponent to get a new exponent of four and divide by this new exponent of four. We get π¦ to the fourth power over four. And weβll do the same for our second term. We add one to the exponent of π¦ to get a new exponent of three and then divide by this new exponent of three. So now weβre subtracting three π¦ cubed minus three. And of course, we can simplify this. Three divided by three is equal to one. So by evaluating this integral, we got π¦ to the fourth power over four minus π¦ cubed. And we still need to evaluate this at the limit of integration π¦ is equal to π and π¦ is equal to zero.

We now want to do the same with our second integral. But we can see our integrand is exactly the same. And if our integrand is exactly the same, we can just choose the same antiderivative: π¦ to the fourth power over four minus π¦ cubed. Then, we just need to evaluate this at the limits of integration, zero and π‘. We now want to evaluate both of these antiderivatives at the limits of integration. However, we can notice something interesting. Both of our antiderivatives evaluated at zero give us zero. So in both cases, we only need to evaluate our antiderivative at one of our limits of integration.

Evaluating our first antiderivative at the lower limit of integration β and remember, because this is the lower limit of integration, we need to subtract this value β we get the limit as π approaches negative β of negative one times π to the fourth power over four minus π cubed. And evaluating our second antiderivative at the upper limit of integration, π¦ is equal to π‘, we get the limit as π‘ approaches β of π‘ to the fourth power over four minus π‘ cubed. Now, remember, for our integral to be convergent, we need both of these limits to converge. Otherwise, we say our integral is divergent. In fact, both of these limits are divergent, and we can show this. Weβll only show one though; weβll look at the second limit.

Thereβs a few different ways of showing this. Weβll take out a factor of π‘ to the fourth power from both terms inside of our limit. This gives us the limit as π‘ approaches β of π‘ to the fourth power times one-quarter minus one over π‘. And now we can evaluate this limit directly. Our limit is as π‘ is approaching β. As π‘ approaches β, π‘ to the fourth power is growing without bound. And of course, one-quarter remains constant. However, one over π‘ is approaching zero as π‘ approaches β. So this entire expression is growing without bound, so this limit is equal to β.

And itβs worth reiterating here when we say that a limit is equal to β, this still means the limit is divergent. So, because one of our limits are divergent, we must have that our original improper integral is also divergent. And this gives us our final answer. Therefore, we were able to show the integral from negative β to β of π¦ cubed minus three π¦ squared with respect to π¦ is divergent.