Limits of sums Use the definition of the defi...

Question

Limits of sums Use the definition of the definite integral to evaluate the following definite integrals. Use right Riemann sums and Theorem 5.1.

∫₀² (2𝓍 + 1) d𝓍

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to use the right remand sums to compute the interval from 1 to 5 of the quantity of 2 X plus 7 in quantity DX using the definition of the definite integral. So, we're gonna need to use the right remand sums as as well as our definition of the definite integral to compute this integral. So, for the remand sums, we first need to define our interval width, which we'll call delta X. And this is going to be equal to. And here, if we look at our limits of integration, we go from 1 to 5, so we'll have the quantity of 5 minus 1 in quantity, and that's going to be divided by N. And when we simplify this expression, we see that delta X is going to be equal to 4 divided by N. Next, we need to find the right endpoints of each of our sub-intervals, and so we'll call those right endpoints X of I. And recall that this is going to be equal to for the right endpoints, or lower limit of integration, which is going to be 1, then we'll have plus I multiplied by delta X, which is going to be 4 divided by N. And here, I is going to take on values from 1 all the way up to N. So, now we need to actually compute our right remand sum. And so we call for the right remand sum, that's going to be the sum from I equal to 1 ton of F of X of I. Multiplied by Delta X. And here F of X is just going to be equal to our integram, which would be 2 X + 7. So making that substitution into this formula, this is going to be equal to the sum. Of I equal to 1 ton. Of the quantity of 2 multiplied by X of I, which is the quantity of 1 plus, or I divided by N. In quantity, then we'll have plus 7 in quantity, multiplied by delta X, which is 4 divided by N. And so multiplying out everything inside of this summation, we'll see that this is going to be equal to the sum of I equal to 1 ton. Of the quantity of 2 + 8 I divided by N plus 7 in quantity multiplied by 4 divided by N. So here we'll collect like terms and distribute the 4 divided by N. In doing so, this is going to be equal to the sum. Of I equal to 1 ton. Of the quantity, and here we'll have 2 + 7, which will give us 9, and we multiply that by 4 divided by N. We'll get 36 divided by N. They will have plus. 32 1 divided by n squared in quantity. So now we want to do is to break this into two separate sums. So this is going to be equal to, and we're going to factor out the constant in front of the summation sign. So we'll have 36 divided by N, multiplied by the sum of I equal to 1 ton of 1. And then we'll have plus 32 divided by N squad, multiplied by the sum of I equal to 1 ton of I. So now we can actually perform the summations. And so doing so, we will get that this is going to be equal to. 36 divided by N multiplied by the sum of I equals 1 ton of 1. I recall that when we perform the sum, we get the value of N. Then we'll have plus 32 divided by n squared, and this gets multiplied by the sum of I equals 1 ton of I. Recall that when we perform this sum, we will get N multiplied by the quantity of N plus 1 in quantity divided by 2. So now we just need to simplify this expression, so this is going to be equal to. For the first term, we'll have 36 divided by n, multiplied by n, which will give you 36. Then we'll have plus for the second term, we will have 32 divided by 2, which will give us 16. Then that gets multiplied by the quantity of n2 plus N N quantity divided by N2. And finally simplifying that last term, this is going to be equal to 36 plus 16, multiplied by the quantity of 1 plus 1 divided by N in quantity. So now that we have calculated our remand sum in terms of N, we can actually use the definition of the definite integral. So that means that our integral here. From 1 to 5 of quantity of 2 X. Plus 7 in quantity, DX. is going to be equal to, and here we call your definition of the definition, using remot sums. That's going to be the limit as N purchase infinity. Of the sum Of I equal to 1 UN of F of X of I multiplied by delta X. Now, we've already calculated to the remand sum, so we'll substitute in that result. So that means that this is going to be equal to the limit. As in approaches infinity. Of the quantity of 36. Plus 16. Multiplied by the quantity of 1 plus 1 divided by N and quantity. And when we take the limit as n approaches infinity, the quantity of 1 divided by N approaches 0. So evaluating the limit, we see that this is going to be equal to 36. Plus 16, multiplied by the quantity of 1 + 0 in quantity. And when we evaluate this result, we see that this is going to be equal to 52. So this here is the answer that we're looking for for this problem. So, I hope that this explanation has been useful, and I'll see you in the next video.

Limits of sums Use the definition of the definite integral to evaluate the following definite integrals. Use right Riemann sums and Theorem 5.1.

∫₀² (2𝓍 + 1) d𝓍

Key Concepts

Definite Integral

Riemann Sums

Theorem 5.1 (Fundamental Theorem of Calculus)

Watch next