Question Video: Estimating the Definite Integration of a Polynomial Function in a Given Interval by Dividing It into Subintervals and Using Their Right Endpoint | Nagwa Question Video: Estimating the Definite Integration of a Polynomial Function in a Given Interval by Dividing It into Subintervals and Using Their Right Endpoint | Nagwa

# Question Video: Estimating the Definite Integration of a Polynomial Function in a Given Interval by Dividing It into Subintervals and Using Their Right Endpoint Mathematics • Higher Education

Approximate the integral β«_(β2) ^(2) (3π₯Β² β 5π₯) dπ₯ using a Riemann sum with right endpoints. Take π to be 8.

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### Video Transcript

Approximate the definite integral between negative two and two of three π₯ squared minus five π₯ with respect to π₯ using a Riemann sum with right endpoints. Take π to be eight.

Remember, we can estimate the solution to a definite integral by using Riemann sums, in this case is the definite integral of three π₯ squared minus five π₯ between the limits of negative two and two. And weβre going to be using right endpoints. When writing Riemann sums with right endpoints, we take values of π from one to π. Itβs the sum of Ξπ₯ times π of π₯π for these values of π, where Ξπ₯ is π minus π divided by π. Bear in mind here that π is the number of subintervals and π₯π is equal to π plus π lots of Ξπ₯. So letβs look at what we actually have.

Our limits are from negative two to two. So weβll let π be equal to negative two and π be equal to two. Weβre told that we need to take π to be eight. Geometrically, this tells us the number of rectangles we have. And now, we can work out Ξπ₯. Thatβs the width of each rectangle. According to our formula, Ξπ₯ is π minus π over π. Thatβs two minus negative two over eight, which is one-half. And once we have Ξπ₯, we can then work out π₯π. Itβs π which we said is negative two plus Ξπ₯. Thatβs half times π. In other words, π₯π is negative two plus π over two.

Now, for our Riemann sum, we need to work out π of π₯π. Thatβs clearly π of negative two plus π over two. We achieve this by substituting negative two plus π over two into our formula three π₯ squared minus five π₯. And if we distribute our parentheses and simplify, we see that π of π₯π is three-quarters π squared minus 17 over two π plus 22. Letβs now substitute Ξπ₯ and π of π₯π into our summation formula. We now see that an approximation to our definite interval is the sum of a half times three-quarters π squared minus 17 over two π plus 22 for values of π from one to eight. Now, actually, this constant factor a half is independent of π. So we could actually take it outside of the sum. And now, whilst this step isnβt entirely necessary, it canβt simplify things on occasion.

Weβre now going to substitute values of π from one through to eight into three-quarters π squared minus 17 over two π plus 22 and find their sum. When π is one, we get 0.75 minus 8.5 plus 22, which is 14.25. When π is two, we get eight. When π is three, we get 3.25. When π is four, we get zero. When π is five, itβs negative 1.75. When π is six, we get negative two. When itβs seven, we get negative 0.75. And when π is eight, we get two. Finding the sum of these values and then multiplying it by one-half and we get 23 over two. And so, an approximation to our definite integral weβre using a right Riemann sum with eight subintervals is 23 over two.

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