The portal has been deactivated. Please contact your portal admin.

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 𝑓(π‘₯) = π‘₯Β³ on the interval [0, 2] in sigma notation using a right Riemann sum with 𝑛 subintervals.

03:54

Video Transcript

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

The question gives us the cubic function 𝑓 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 𝑛 subintervals. Let’s 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 𝑓 of π‘₯ on the closed interval from π‘Ž to 𝑏 by calculating the sum from 𝑖 equals one to 𝑛 of Ξ”π‘₯ times 𝑓 evaluated at π‘₯ 𝑖, where Ξ”π‘₯ is the width of our subinterval. So if we have 𝑛 subintervals, this will be equal to 𝑏 minus π‘Ž all divided by 𝑛. And our π‘₯ 𝑖 will be the sample points. Since we’re taking a right Riemann sum, these will be the right endpoints for our subintervals. π‘₯ 𝑖 will be equal to π‘Ž plus Ξ”π‘₯ times 𝑖.

Since we want to represent the area under the curve of the function 𝑓 of π‘₯ is equal to π‘₯ cubed on the closed interval from zero to two, we set our function 𝑓 of π‘₯ to be π‘₯ cubed, our value of π‘Ž equal to zero, and our value of 𝑏 equal to two. We can now find Ξ”π‘₯. It’s equal to 𝑏 minus π‘Ž divided by 𝑛. In our case, π‘Ž is equal to zero and 𝑏 is equal to two. And remember, we’re just told to use 𝑛 subintervals. So we get Ξ”π‘₯ is equal to two minus zero all divided by 𝑛. And of course we can simplify this to get Ξ”π‘₯ is equal to two over 𝑛.

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

We’re 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 𝑖 equals one to 𝑛 of Ξ”π‘₯ times 𝑓 evaluated at π‘₯ 𝑖. We showed that Ξ”π‘₯ is equal to two over 𝑛 and π‘₯ 𝑖 is equal to two 𝑖 over 𝑛. So by substituting these values in and remembering that our function 𝑓 is equal to π‘₯ cubed, we now have the sum from 𝑖 equals one to 𝑛 of two over 𝑛 times two 𝑖 over 𝑛 all cubed.

The next thing we’ll do is cube two 𝑖 over 𝑛. Doing this, we get two cubed times 𝑖 cubed divided by 𝑛 cubed. And of course two cubed is equal to eight. So we now have the sum from 𝑖 equals one to 𝑛 of two over 𝑛 times eight 𝑖 cubed divided by 𝑛 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 𝑛 times 𝑛 cubed to give us 𝑛 to the fourth power. So we now have the sum from 𝑖 equals one to 𝑛 of 16𝑖 cubed divided by 𝑛 to the fourth power.

Now, we could leave our answer like this. However, remember, our sum is as 𝑖 is going from one to 𝑛. Our value of 𝑛 remains constant in this sum. This means we can treat it as a constant. So, in fact, in our summand, 16 divided by 𝑛 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 𝑛 to the fourth power multiplied by the sum from 𝑖 equals one to 𝑛 of 𝑖 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 𝑓 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 𝑛 subintervals. We got that this area was approximately equal to 16 divided by 𝑛 to the fourth power multiplied by the sum from 𝑖 equals one to 𝑛 of 𝑖 cubed.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.