Integration by Riemann sums Consider the inte...

Question

Integration by Riemann sums Consider the integral ∫₁⁴ (3𝓍― 2) d𝓍.

(a) Evaluate the right Riemann sum for the integral with n = 3 .

Accepted Answer

Hi everyone. Let's take a look at this practice problem. This problem says compute the right Riemann sum or the interval from -1 to 2 of the quantity of 5 X plus 4 and quantity DX using N equal to 3 subintervals. And we're given 4 possible choices as our answers. For choice A, we have 20, for choice B, we have 22. Choice C, we have 12, and for choice D, we have 27. So, the first thing we want to do is to figure out our interval width. So, that is going to be Delta X. I recall that this is going to be equal to the quantity of B minus A in quantity divided by N. Where B here is the upper bound of our limits of integration, A is going to be the lower bound for the limits of integration, and N is the number of sub-intervals. So, in our case, Delta X is going to be equal to the quantity of 2, since that is the upper bound, minus, and for the lower bound, we'll have -1. And that quantity is going to be divided by 3 since N is equal to 3, since we're looking for 3 sub-intervals. And when we evaluate this expression, we see that Delta X is going to be equal to 1. So now we can actually determine our sub-intervals. So, our first sub interval is going to be the interval from -1 to 0. The next interval is going to be from 0 to 1. And our final interval is going to be from 1 to 2. And here we're going to compute the right Riemann sum. So we want to use the right endpoint of each of the intervals. So that means that the right end point for the first interval, which we'll call X1, is going to be equal to 0. You write end point for the second info, which we'll call X2, it's going to be equal to 1. And the right end point for the 3rd inval, which we'll call X3, is going to be equal to 2. So now we'll calculate the right remand sum, and recall that for the right remand sum, we're going to have the sum. From K equal to 1. 23. Of the quantity of F of X of K. Multiplied by Delta X. Which is going to be equal to the quantity of F of X1. Plus F of X2. Plus F of X3. In quantity multiplied by delta X. So here we actually need to calculate the F of X1, F of X2, and F of X3. Now, in our case, F of X, It's just gonna be our integrand, which is going to be equal to 5 X plus 4. That means that F of X1 is going to be equal to F of 0, which is equal to 5 multiplied by 0 + 4, which is equal to 4. Or F of X2. This is going to be equal to F of 1. Which is equal to 5 multiplied by 1 plus 4, which is equal to 9. And F of X3. Is going to be equal to F of 2, which is equal to 5 multiplied by 2. Plus 4, which is equal to 14. So, substituting in these values into our expression for the Riemann sum, we see that that is going to be equal to the quantity of 4. Plus 9 Plus 14 in quantity multiplied by delta X, which is just 1. And when we evaluate this expression, we see that this is going to be equal to. 27. So that is the answer that we're looking for for this problem. And when we compare our answer to our answer choices, we see that the correct answer choice corresponds to answer choice D. So, I hope that this explanation has been useful, and I'll see you in the next video.

Integration by Riemann sums Consider the integral ∫₁⁴ (3𝓍― 2) d𝓍.

(a) Evaluate the right Riemann sum for the integral with n = 3 .

Key Concepts

Riemann Sums

Definite Integral

Partitioning the Interval

Watch next