Can Anyone help with this question pls!!!?

[takelight]

In this problem you will calculate ∫3xdx by using the formal definition of the definite integral:

∫(b) (a) f(x)dx=lim(n→∞) [∑(n) (k=1) f(x_k^* ) Δx].

(a) The interval [1,5] is divided into n equal subintervals of length Δx . What is Δx (in terms of n )?
Δx = 4/n (I GOT THIS ONE RIGHT)
(b) The right-hand endpoint of the kth subinterval is denoted x_k^* . What is x_k^* (in terms of k and n )?
I'm stuck on this one.. I tried 4k/n but it's wrong. I really don't know this one.

 

[Dr.Peterson]

Start by writing simple expressions for the 1st, 2nd, 3rd right-hand endpoints. See if that suggests a general formula. Show us your expressions.

The left-hand endpoint of the first interval is 1. What is the right-hand endpoint of that interval?

 

[takelight]

Start by writing simple expressions for the 1st, 2nd, 3rd right-hand endpoints. See if that suggests a general formula. Show us your expressions.

The left-hand endpoint of the first interval is 1. What is the right-hand endpoint of that interval?

Um. the first right hand endpoint would be the length of the interval times the place where the interval is at. That's what I think at least, which is why I did 4k/n... I'm not sure I am able to solve this part.

 

[JeffM]

[MATH]\Delta x = \dfrac{5 - 1}{n} = \dfrac{4}{n}.[/MATH] You got this right. Great.

What Dr. Peterson asked you to do is to write this out.

The x-value at the first right endpoint is [MATH]1 + \Delta x = 1 +\left (1 * \dfrac{4}{n} \right ).[/MATH]
Of course, that means it is also the starting x-value of the second interval so

the x-value at the second right endpoint is [MATH](1 + \Delta x) + \Delta x = 1 + \left (2 * \dfrac{4}{n} \right ).[/MATH]
And of course the right end-point of the n^th interval (the last interval) is

[MATH]5 = 1 + 4 = 1 + \left ( n * \dfrac{4}{n} \right ).[/MATH]
So if we pick the k^th interval, what do you think the x-value of the right endpoint might be?

What would be the y-value there?

 

[Dr.Peterson]

Um. the first right hand endpoint would be the length of the interval times the place where the interval is at. That's what I think at least, which is why I did 4k/n... I'm not sure I am able to solve this part.

So what IS that first endpoint? I want to see your actual answer. What you describe doesn't sound right.

Take it bit by bit as I suggested, and you'll get to the answer. Don't rush.

In fact, let's back up even more. What are the right-hand endpoints if n=4, so that Δx = 4/4 = 1, specifically?

 

[takelight]

So what IS that first endpoint? I want to see your actual answer. What you describe doesn't sound right.

Take it bit by bit as I suggested, and you'll get to the answer. Don't rush.

In fact, let's back up even more. What are the right-hand endpoints if n=4, so that Δx = 4/4 = 1, specifically?

Oh I get it. So it's 1+ 4k/n because we start at 1 and take k steps of Delta x length to reach the right endpoint of the kth interval. And since delta x is 4/n, It checks out.

 

[Dr.Peterson]

That's it: You start at the left-hand end of the interval, and add Δx for each subinterval. Since the first one is after the first subinterval, you've added k times 4/n at the kth subinterval. And the nth point is 1 + n*4/n = 1 + 4 = 5, as needed.

What would be different for left-hand endpoints?