Definite Integrals

In Exercises 5–8, express ...

Question

Definite Integrals

In Exercises 5–8, express each limit as a definite integral. Then evaluate the integral to find the value of the limit. In each case, P is a partition of the given interval, and the numbers cₖ are chosen from the subintervals of P.

n

lim ∑ (2cₖ - 1)⁻¹/² ∆xₖ, where P is a partition of [1, 5]

∥P∥→0 k = 1

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Evaluate the limit by first expressing it as a definite integral. The limit. As the norm of P approaches zero of the sum evaluated from K equals 1 ton of C subscript K + 1. To the power of 1/2 multiplied by delta X subscript K. Where P is a partition of square brackets of 0.8 and C subscript K is a sample point chosen within the kth subinterval of the partition. OK. So it appears for this particular prompt, we're ultimately asked to evaluate this limit by expressing it first as a definite integral, and we are asked to do this with all the information that is provided to us by the prophet itself. So by taking all the information that is provided to us, we're asked to use this information to help us evaluate the limit. And in order to do that, we will need to first express it as a definite integral. So with that in mind, we need to be able to recall and note that the given limit by the prom itself is a form of Riemann sum for the definite integral of a function. F of X over the closed interval square brackets of 0.8 is square brackets indicate a closed interval, and we need to note that this is where, so once again, where, let's make a note, where F of X, so this will be where. F of X will be equal to drum rolls of x + 1 to the power of 1/2. So once again, we need to be able to recognize that the limit that is given to us by the prom itself is a form of Riemann sum or a definite integral of the function f of x. Over the closed interval 0.8, where F of X is equal to x + 1 parentheses to the power of 1/2. And we need to note that as the norm of P, so as the norm of P approaches 0, that will mean that the Riemann sum, which is the sum evaluated from K is equal to 1. Ton of C subscript K, which I'll just call CK for short, so the sum evaluated from k equals 1 ton of CK plus 1 in parentheses to the power of 1/2 multiplied by delta XK. We're X subscript KXK for short. And we need to note that this will converge to the definite integral, so I'll draw an arrow to indicate that it will converge to the integral evaluated from 0 to 8 of x + 1 in parentheses to the power of 1/2 dx. OK, we are on a roll. So now, at this point, we need to compute this definite integral, and in order to do that, as we should recall, we need to find the antiderivative. So, with that said, let us state that you will be equal to x + 1. So that will mean that DU will be equal to DX. Then that will mean. That we could go ahead and write that the integral. Of X + 1 in parentheses to the power of 1/2. DX will be equal to the integral of u to the power of 1/2 DU, which will be equal to 2 multiplied by u to the power of 1/2. Plusc where c denotes some constant interval. And this will be equal to 2 multiplied by. X + 1 in parentheses to the power of 1/2. Plus C, fantastic. So now we need to evaluate our definite integral. So we need to take the integral, evaluated from 0 to 8. Of x + 1 in parentheses to the power of 1/2. D x, which will be equal to 2 multiplied by x plus 1 in parentheses to the power of 1/2. Evaluated from 0 to 8. Which will be Equal to. 2 multiplied by. 8 + 1 parentheses the power of 1/2 minus 2 multiplied by 0 + 1 parentheses to the power of 1/2. Which will be equal to 2 multiplied by the square root of 9 minus 2 multiplied by the square root of 1, which when we perform that when we perform that math in our calculator, so when we plug this expression into our calculator, we will find that it's simply equal to 4 and 4. Is our correct and final answer. Therefore, without a doubt, the value of our limit is 4. So to quickly recap what we've learned, our final answer without a doubt will be simply 4, and that is our evaluated limit, and that's it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

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