### Video Transcript

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

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

By using a right Riemann sum, we can approximate the area under the curve of a function π of π₯ on the closed interval from π to π with π subintervals as the sum from π equals one to π of Ξπ₯ times π evaluated at π₯ π, where Ξπ₯ will be our subinterval width. This will be the length of our interval divided by the number of subintervals. So we have Ξπ₯ is equal to π minus π divided by π. And since weβre taking a right Riemann sum, we want our sample points to be the right endpoints of our subintervals. In this case, π₯ π will be equal to π plus Ξπ₯ times π.

Since we want to estimate the area under the curve of the function π of π₯ is equal to π₯ squared plus two π₯ plus one on the closed interval from zero to three by using our Riemann sum, weβll set our function π of π₯ to be π₯ squared plus two π₯ plus one, our value of π to be zero, and our value of π to be equal to three.

Weβre now ready to find the value of Ξπ₯. Remember, Ξπ₯ will be π minus π divided by the number of subintervals π. In our question, weβre told to use π subintervals. So we get Ξπ₯ is equal to three minus zero divided by π. And we can simplify this to just be three divided by π.

Now that we found the value of Ξπ₯, we can find an expression for each of our sample points π₯ π. So by using the fact that π is equal to zero and Ξπ₯ is equal to three over π, we get that π₯ π is equal to zero plus three over π times π. And weβll simplify this to get three π divided by π.

Now that we found expressions for Ξπ₯ and π₯ π and we know our function π of π₯, we can use these in our Riemann sum to approximate the area under our curve. Substituting in our expressions for Ξπ₯ and π₯ π, we get the sum from π equals one to π of three over π times π evaluated at three π divided by π. Remember, our function π of π₯ is π₯ squared plus two π₯ plus one. So we need to substitute this expression for π₯ into our function. Doing this, we get the sum from π equals one to π of three over π times three π over π all squared plus two times three π over π plus one.

We could leave our answer like this. However, our series has π from one to π, so our value of π is varying. However, inside of our series, our value of π is not varying; itβs a constant. So we can take it outside of our series. So we can use this to simplify our series. Weβll start by taking the constant factor of three over π outside of our series. So this gives us three over π times the sum from π equals one to π of three π over π all squared plus two times three π over π plus one.

Letβs now simplify our summand. Weβll start by distributing the square over the first term in our summand. This gives us nine π squared divided by π squared. Next, we can simplify the second term in our summand since two times three π divided by π is six π divided by π. So our new summand is nine π squared divided by π squared plus six π divided by π plus one.

And thereβs one more piece of simplification we can do. Remember, our value of π inside of this series is a constant. So we could simplify our summand by multiplying it by π squared. Of course then our answer will be wrong by a factor of π squared. So weβll just divide our entire answer by π squared. This gives us the following expression, and we can simplify. One over π squared times three over π is equal to three over π cubed. So the coefficient of our series is three over π cubed.

Next, weβll distribute π squared over our parentheses and then simplify each term. Doing this, we get a new summand of nine π squared plus six ππ plus π squared. And since π is a constant, weβll rearrange our second term inside of our summand to be six ππ. And this gives us our final answer.

Therefore, we were able to approximate the area under the curve of the function π of π₯ is equal to π₯ squared plus two π₯ plus one on the closed interval from zero to three by using a right Riemann sum with π subintervals. We got this area was approximately equal to three over π cubed times the sum from π equals one to π of nine π squared plus six ππ plus π squared.