### Video Transcript

Express the limit as π approaches β of one over π times the sum from π₯ equals one to π of five divided by four minus π₯ over π all squared as a definite integral.

In this question, we need to express a limit as a definite integral. And we can see that this limit should remind us of a Riemann sum. So to answer this question, weβll start by recalling what we know about writing definite integrals as limits of Riemann sums.

We recall if π is an integrable function on a closed interval from π to π, then the definite integral from π to π of π of π₯ with respect to π₯ will be equal to the limit as π approaches β of the right Riemann sum of π of π₯. This is the limit as π approaches β of the sum from π equals one to π of π of π₯π times Ξπ₯, where Ξπ₯ are the widths of our intervals, π minus π divided by π. And π₯π will be our sample points, π plus π times Ξπ₯.

And itβs worth pointing out we donβt actually need to use the right Riemann sum. We can use any of the Riemann sums. Itβs just standard to use the right Riemann sums. And, in fact, in this question, we can see our sum goes from π₯ is equal to one to π. So we would need to use right Riemann sums to answer this question anyway.

So to answer this question, weβre going to need to start by rewriting the limit given to us in the question in the following form. And, in fact, we can see itβs almost in this form already. Thereβs just two main problems. First, our sum is over π₯ instead of π. And second, weβre multiplying our entire sum by one over π. In fact, we can fix both of these problems.

First, instead of using π₯, weβll just use π₯ is equal to π. Next, we need to notice that one over π is a constant with respect to our series. It doesnβt change as the value of π₯ changes. So we can just bring this inside of our series.

So by using both of these changes, weβve rewritten the limit given to us in the question as the limit as π approaches β of the sum from π equals one to π of five divided by four minus π over π all squared all multiplied by one over π. And now we can see this is very similar to the limit as π approaches β of a right Riemann sum. Weβll set our value of Ξπ₯ to be one divided by π. And we know that Ξπ₯ should be equal to π minus π over π.

So if we want Ξπ₯ to be one over π, this should be equal to π minus π divided by π. And we can multiply through by π. This gives us one is equal to π minus π. And since weβre trying Ξπ₯ is equal to one over π, we can substitute this into our expression for π₯π to find an expression for our sample points.

Substituting this in, we get π₯π is equal to π plus π times one over π. And, of course, we can simplify this to give us π plus π divided by π. Remember, in a right Riemann sum, we evaluate π at each of our sample points π₯π. So we would need this to appear in our limit. And, in fact, we can see something interesting. We have something which is very similar to π₯π appearing in this expression. We have π divided by π. This is in fact π₯π if our value of π is equal to zero. So letβs start by choosing π is equal to zero.

Remember, we already showed that one is equal to π minus π. So if π is zero, we must have that π is equal to one. So weβll set π equal to zero and π equal to one. And then with π equal to zero and π equal to one, we can find Ξπ₯ and π₯π for our right Riemann sum. We get Ξπ₯ is one over π. And our sample points π₯π will be π divided by π.

The last thing weβre going to need to write this as a right Riemann sum is to find an expression for π of π₯. Remember, five over four minus π over π all squared should be π evaluated at π₯π. And we can see that π₯π directly appears in this expression. So we just need to set π of π₯ equal to five divided by four minus π₯ squared. Then this is just π evaluated at π₯π.

So now weβre ready to construct a right Riemann sum with the following parameters. π is zero, π is one, and π of π₯ is five divided by four minus π₯ squared. Using this, we can rewrite the limit given to us in the question as the limit as π approaches β of the sum from π equals one to π of π of π₯π times Ξπ₯. This is the limit as π approaches β of the right Riemann sum of five divided by four minus π₯ squared between the limits of integration zero and one.

Now, we might be tempted to just write this as the integral from zero to one of our function π of π₯ with respect to π₯. But remember, we can only do this when π is integrable on the closed interval from π to π. So we need to check this is true in our case.

First, remember, our function π of π₯ is five divided by four minus π₯ squared. Next, our value of π is zero and π is one. So we need to check that π is integrable on the closed interval from zero to one. One way to show that π is integrable on the closed interval from zero to one is to notice that π is a rational function. And we know rational functions are continuous across their entire domain. And if a function is continuous on an interval, then it must be integrable on that interval.

So, in fact, all we need to do is check that our function π of π₯ is defined for all values of π₯ on the closed interval from zero to one. And we also know something interesting about rational functions. Theyβll be defined everywhere except when their dominator is equal to zero. So we just need to check where the denominator of this rational function is equal to zero. It will be equal to zero when π₯ is equal to positive or negative two. And, of course, these two values of π₯ are not in our interval. So our function π of π₯ is integrable on the closed interval from zero to one. And this justifies the use of our rule.

We know that the limit of the right Riemann sum as π approaches β will in fact be equal to the definite integral. So we can just write our values of π, π, and π of π₯ into the definite integral. This gives us the integral from zero to one of five divided by four minus π₯ squared with respect to π₯.

In this question, we were given the limit and we needed to write this as a definite integral. We were able to rewrite this limit as a right Riemann sum and then proved that this right Riemann sum must converge to the definite integral from zero to one of five divided by four minus π₯ squared with respect to π₯.