# Question Video: Representing the Area Under a Given Section of a Curve Using a Riemann Sum Mathematics • Higher Education

Represent the area under the curve of the function π(π₯) = π₯Β² + 2π₯ + 1 on the interval [0, 3] in sigma notation using a right Riemann sum with π subintervals.

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### 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. Letβs 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 Ξπ₯ times π evaluated at π₯ π, where Ξπ₯ will be our subinterval width. This will be the length of our interval divided by the number of subintervals. So we have Ξπ₯ is equal to π minus π divided by π. And since weβre 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 Ξπ₯ 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, weβll 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.

Weβre now ready to find the value of Ξπ₯. Remember, Ξπ₯ will be π minus π divided by the number of subintervals π. In our question, weβre told to use π subintervals. So we get Ξπ₯ 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 Ξπ₯, 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 Ξπ₯ is equal to three over π, we get that π₯ π is equal to zero plus three over π times π. And weβll simplify this to get three π divided by π.

Now that we found expressions for Ξπ₯ 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 Ξπ₯ 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; itβs a constant. So we can take it outside of our series. So we can use this to simplify our series. Weβll 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.

Letβs now simplify our summand. Weβll 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 thereβs 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 weβll 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, weβll distribute π squared over our parentheses and then simplify each term. Doing this, we get a new summand of nine π squared plus six ππ plus π squared. And since π is a constant, weβll rearrange our second term inside of our summand to be six ππ. 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 ππ plus π squared.