### Video Transcript

Represent the area under the curve of the function π of π₯ is equal to π₯ cubed on the closed interval from zero to two in sigma notation using a right Riemann sum with π subintervals.

The question gives us the cubic function π of π₯. It wants us to represent the area under the curve of this function on the closed interval from zero to two in sigma notation by using a right Riemann sum with π subintervals. Letβs start by recalling what we mean by a right Riemann sum.

A right Riemann sum tells us we can approximate the area under the curve of the function π of π₯ on the closed interval from π to π by calculating the sum from π equals one to π of Ξπ₯ times π evaluated at π₯ π, where Ξπ₯ is the width of our subinterval. So if we have π subintervals, this will be equal to π minus π all divided by π. And our π₯ π will be the sample points. Since weβre taking a right Riemann sum, these will be the right endpoints for our subintervals. π₯ π will be equal to π plus Ξπ₯ times π.

Since we want to represent the area under the curve of the function π of π₯ is equal to π₯ cubed on the closed interval from zero to two, we set our function π of π₯ to be π₯ cubed, our value of π equal to zero, and our value of π equal to two. We can now find Ξπ₯. Itβs equal to π minus π divided by π. In our case, π is equal to zero and π is equal to two. And remember, weβre just told to use π subintervals. So we get Ξπ₯ is equal to two minus zero all divided by π. And of course we can simplify this to get Ξπ₯ is equal to two over π.

Now that we found Ξπ₯, we have all the information we need to find an expression for each of our sample points π₯ π. Substituting in our value of π equal to zero and Ξπ₯ equal to two over π, we get that π₯ π is equal to zero plus two over π times π. And of course we can simplify this to get that π₯ π is equal to two π divided by π.

Weβre now ready to find an expression for our area. By using our right Riemann sum, we have this area is approximately equal to the sum from π equals one to π of Ξπ₯ times π evaluated at π₯ π. We showed that Ξπ₯ is equal to two over π and π₯ π is equal to two π over π. So by substituting these values in and remembering that our function π is equal to π₯ cubed, we now have the sum from π equals one to π of two over π times two π over π all cubed.

The next thing weβll do is cube two π over π. Doing this, we get two cubed times π cubed divided by π cubed. And of course two cubed is equal to eight. So we now have the sum from π equals one to π of two over π times eight π cubed divided by π cubed.

We now want to simplify this expression by multiplying the two factors in our summand together. In our numerator, we can simplify two times eight to give us 16. And in our denominator, we can simplify π times π cubed to give us π to the fourth power. So we now have the sum from π equals one to π of 16π cubed divided by π to the fourth power.

Now, we could leave our answer like this. However, remember, our sum is as π is going from one to π. Our value of π remains constant in this sum. This means we can treat it as a constant. So, in fact, in our summand, 16 divided by π to the fourth power is a constant. So we can take this outside of our sum. And so by taking this constant factor outside of our series, we get 16 divided by π to the fourth power multiplied by the sum from π equals one to π of π cubed. And we can see this is in sigma notation. So we can leave our answer like this.

Therefore, we were able to represent the area under the curve of the function π of π₯ is equal to π₯ cubed on the closed interval from zero to two in sigma notation by using a right Riemann sum with π subintervals. We got that this area was approximately equal to 16 divided by π to the fourth power multiplied by the sum from π equals one to π of π cubed.