### Video Transcript

Determine whether the integral from negative two to three of one over π₯ to the fourth power with respect to π₯ is convergent or divergent.

Letβs begin by graphing this function. Looking at this graph, we can see that this function is undefined at π₯ equals zero. Thatβs because we can see a vertical asymptote at π₯ equals zero. So, the function is discontinuous at π₯ equals zero, which means that this integral is called an improper integral. Fortunately, we have a rule which helps us work out whether an improper integral is convergent or divergent. That is, the improper integral between π and π of π of π₯ with respect to π₯ is called convergent if the corresponding limit exists and divergent if the limit does not exist.

And we deal with these kind of integrals by splitting the integral up. If π has a discontinuity at π which is between π and π, then we define the integral between π and π to be the integral between π and π of that function add the integral between π and π of that function. So, weβre basically splitting the integral at the point of the discontinuity.

So, because for our function the discontinuity is at π₯ equals zero, we split our integral into the integral between negative two and zero of one over π₯ to the fourth power with respect to π₯ add the integral between zero and three of one over π₯ to the fourth power with respect to π₯.

Letβs firstly evaluate this integral here. We can start by writing this as the negative exponent, π₯ to the power of negative four. And we deal with this kind of integral by replacing the discontinuous limit with a variable and then taking the limit of the integral as that variable approaches the discontinuous limit.

So, if we replace the zero limit with π‘ and then take the limit as π‘ approaches zero from the left of the integral between negative two and π‘ of π₯ to the power of negative four with respect to π₯. We can then integrate π₯ to the power of negative four with our usual integration rule of adding one to the power and then dividing by the new power.

This gives us negative π₯ to the power of negative three over three. And seeing as we have limits, we donβt need to include a constant of integration. Before we substitute in our limits, letβs simplify negative π₯ to the power of negative three over three. This is negative one over π₯ cubed over three, which we can write as negative one over three π₯ cubed. So now, letβs substitute in our limits.

We have to be careful here remembering that there was a negative sign before our function. So, weβre going to be subtracting a negative when we substitute in the limits. If we take the first term in our bracket here, negative one over three π‘ cubed, and we take the limit as π‘ approaches zero from the left, we can see that this will make the denominator zero. So, this term will approach infinity. So, the limit of the whole expression will be infinity. So, that gives us the value of the first integral.

And because this limit doesnβt exist, it doesnβt go to a specific value, we can say that that integral is divergent. And actually, we donβt need to go any further with this question because if either part of the integral diverges, then the integral itself diverges. So, we can conclude that this integral is divergent.