Integration by Riemann sums Consider the inte...

Question

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

(c) Evaluate the definite integral by taking the limit as n →∞ of the Riemann sum in part (b).

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to find the value of the integral from 0 to 3 of the quantity of 2 X plus 5 in quantity DX by expressing it as the limit of a remand sum using right endpoints. And we're given 4 possible choices as our answers. For choice A, we have 14, for choice B, we have 12. For choice C, we have 24, and for choice D, we have 48. So, the first step into solving this problem, we need to first find our interval width for a rebound sum. That is going to be delta X. And recall the delta X is going to be equal to the quantity of the upper limit of our limits of integration, which is going to be 3, minus the lower limit, which is 0, and then that quantity is going to be divided by N. And so when we simplify this expression, we see that delta X is going to be equal to 3 divided by N. Next thing we need to find the values of X for each of our right endpoints or our sub-intervals in our remand sum. So, we'll label those endpoints as X of I, and for the right endpoints, this is going to be equal to the lower bound. Of our limits of integration, which is going to be 0. Plus I multiplied by delta X, which in this case is 3 divided by N. And so find this expression, we see that this is going to be equal to 3 I divided by N. where I here gonna be going from 1 all the way up to N. So now we can perform our remod sum. So we call that for the remod sum, we'll have the sum. Of I equal to 12 N. Of F of X of I. Multiplied by Delta X. So this is going to be equal to. The sum Of I equal to 12N. Of the quantity, and here for F of X, we'll just use our integrand in our integral. So F of X is equal to 2 X plus 5, so we'll have 2 multiplied by X of I, which is the quantity of 3 I divided by N. In quantity, it will have less 5. In quantity multiplied by delta X, which is 3 divided by N. So simplifying this expression, this is going to be equal to the sum of I equal to 1 ton. Of the quantity, and here multiplying everything out, we'll get 18, multiplied by I divided by N2. Plus 15 divided by N. In quantity So, now rewriting this expression as two separate sums and factoring out the consonants in front of the summation sign. This is going to be equal to 18 divided by N2d, multiplied by the sum of I equal to 1 ton of I. Plus 15 divided by N. Multiplied by the sum of I equal to 1 ton of 1. Which is going to be equal to. 18 divided by n2d and recall that the sum of I equals 1 ton of I is going to be equal to the quantity of N multiplied by the quantity of N plus 1 in quantity divided by 2. In quantity. Then we'll have less. 15 divided by N. Multiplied by, and here the sum of I equal 1 ton of 1 is just equal to n. So again, we just need to simplify this expression, so this is going to be equal to. 18 divided by N2d. Multiplied by the quantity of N squared. Plus N, and that quantity is going to be divided by 2, and quantity, plus 15. And continuing to simplify, you see that this is going to be equal to. 9 multiplied by the quantity of 1 plus 1 divided by N in quantity plus 15. And so, in order to find the value of the integral, we need to take the limit of our Riemann sum as N approaches infinity. And so the limit As in purchases infinity. Of ariemansson, which is the sum of I equal to 1 UN of F of X of I. Multiplied by Delta X, that this is going to be equal to the limit. As in approaches infinity. Of the quantity of 9 multiplied by the quantity of 1 plus 1 divided by N in quantity plus 15. Inquiti And when we take the limit as n approaches infinity, the quantity of 1 divided by n approaches 0. So this is going to be equal to 9 multiplied by the quantity of 1 + 0 in quantity plus 15, which when we evaluate this expression is going to be equal to 24. So this here is the answer that we're looking for for this problem. And when we compare that to our answer choices, we see that the correct answer choice corresponds to answer choice C. 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.

(c) Evaluate the definite integral by taking the limit as n →∞ of the Riemann sum in part (b).

Key Concepts

Riemann Sums

Definite Integral

Limit Process

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