Solving Integration Help Homework: Lim inf x->inf Ʃ e^(sqrt ti)/N(sqrt ti)

[smiggles]

Homework Statement

The interval [0,4] is partitioned into n equal subintervals and ti, is chosen in [Xi-1, Xi] for each i.
What is:

Lim... inf
x->inf...Ʃ e^(sqrt ti)/N(sqrt ti)
...i=1

The Attempt at a Solution

It was problem we had to do in class as a group, teacher wasnt there to provide us an answer, and some people in my group have conflicting answersWhat i did:

I wrote it as integral

4
∫ (e^(sqrt X))/(N(sqrt X))
0

I then use the substitution method:
U= sqrt X
du= 1/(2 sqrtX) dx
2du= 1/(sqrt X)

...4
2/n ∫ e^u = (2e^(sqrt 4))-(2e/n^(sqrt 0))
...0

= 2e^2-2
My answer = 2/n(e^2 - 1)
Can someone help me out, i don't really understand what we are suppose to do, and i apologize for the bad input. Appreciate it.

 

[SammyS]

Homework Statement

The interval [0,4] is partitioned into n equal sub-intervals and ti, is chosen in [Xi-1, Xi] for each i.
What is:

Lim... inf
x->inf...Ʃ e^(sqrt ti)/N(sqrt ti)
...i=1

The Attempt at a Solution

It was problem we had to do in class as a group, teacher wasnt there to provide us an answer, and some people in my group have conflicting answers


What i did:

I wrote it as integral

4
∫ (e^(sqrt X))/(N(sqrt X))
0

I then use the substitution method:
U= sqrt X
du= 1/(2 sqrtX) dx
2du= 1/(sqrt X)

...4
2/n ∫ e^u = (2e^(sqrt 4))-(2e/n^(sqrt 0))
...0

= 2e^2-2
My answer = 2/n(e^2 - 1)

Can someone help me out, i don't really understand what we are suppose to do, and i apologize for the bad input. Appreciate it.

Hello smiggles. Welcome to PF !

I assume that N and n are the same thing.

What quantity represents Δx in the sum ?

 

[smiggles]

Hello smiggles. Welcome to PF !

I assume that N and n are the same thing.

What quantity represents Δx in the sum ?

Hey sammy, thanks for the reply!

Yes N and n are the same thing.

im not sure if i under stand what youre asking me, but would Δx = (b-a)/n = (4-0)/n = 4/n

 

[SammyS]

Hey sammy, thanks for the reply!

Yes N and n are the same thing.

I'm not sure if i under stand what you're asking me, but would Δx = (b-a)/n = (4-0)/n = 4/n

Yes, Δx = 4/n .

Your integral does not have a dx in it. The quantity in the sum that would correspond to dx is Δx . Can you figure out how to have Δx, i.e. 4/n, in your sum?

 

[smiggles]

4
∫ (e^(sqrt x))/(n(sqrt x)) * (4/n)
0

or could i just multiply the (4/n) to my answer:

= (8/n^2)(e^2 - 1)

 

[SammyS]

4
∫ (e^(sqrt x))/(n(sqrt x)) * (4/n)
0

or could i just multiply the (4/n) to my answer:

= (8/n^2)(e^2 - 1)

There's still no dx there ... but that's not what I asked you to do anyway.

How can you get 4/n to be in the sum ?

\displaystyle \sum_{i=1}^{\infty} \frac{e^\sqrt{t_i}}{n\sqrt{t_i}}

 

[smiggles]

dx= 4/n

can i solve for n, so it would be n=(dx/4), and i would plug it in?
:(

edit:

or can i just tack on a dx
\displaystyle \sum_{i=1}^{\infty} \frac{e^\sqrt{t_i}}{n\sqrt{t_i}} DX

2nd edit:
to get 4/n to be in the sum, can i put it in front of the sigma?

4/n\displaystyle \sum_{i=1}^{\infty} \frac{e^\sqrt{t_i}}{n\sqrt{t_i}}

 

[SammyS]

dx= 4/n

can i solve for n, so it would be n=(dx/4), and i would plug it in?
:(

edit:

or can i just tack on a dx
\displaystyle \sum_{i=1}^{\infty} \frac{e^\sqrt{t_i}}{n\sqrt{t_i}} DX

2nd edit:
to get 4/n to be in the sum, can i put it in front of the sigma?

4/n\displaystyle \sum_{i=1}^{\infty} \frac{e^\sqrt{t_i}}{n\sqrt{t_i}}

No. It's like this:

\displaystyle \sum_{i=1}^{\infty} \frac{e^\sqrt{t_i}}{n\sqrt{t_i}}=\frac{1}{4}\sum_{i=1}^{\infty} \frac{4e^\sqrt{t_i}}{n\sqrt{t_i}}

By the way, maybe we should be looking at \displaystyle \lim_{n\to\infty} \sum_{i=1}^{n} \frac{e^\sqrt{t_i}}{n\sqrt{t_i}}\ .

 

[smiggles]

Nice, one question, where did the 1/4 come from?

Looking back at \displaystyle \lim_{n\to\infty} \sum_{i=1}^{n} \frac{e^\sqrt{t_i}}{n\sqrt{t_i}}\ .

My first instict would be to change
\frac{e^\sqrt{t_i}}{n\sqrt{t_i}}\ .
to

4
∫ fx dx = \frac{e^\sqrt{x}}{n\sqrt{x}}\ dx .
0

Is there something I am overlooking, or a step I am skipping?

I really appreciate your help, and your great patience!

 

[SammyS]

The use of t_i instead of some x_i (I know, it's because x_i has a different use.) may be making this a little less straight forward.

You need something like:

\displaystyle \sum_{i=1}^{n} (f(t_i)\Delta x)

to become

\displaystyle \int_{a}^{b} f(x)\,dx\ .

The sum \displaystyle \ \ \frac{1}{4}\sum_{ i=1}^{\infty} \frac{4e^\sqrt{t_i}}{n\sqrt{t_i}}\ \ does that.

\displaystyle \frac{e^\sqrt{t_i}}{\sqrt{t_i}} is f(t_i).

\displaystyle \frac{4}{n}\ \ is Δx .

 

[smiggles]

so i should have
1/4\displaystyle \\\int_{0}^{4} \ \frac{4e^\sqrt{x}}{n\sqrt{x}}\ \

sorry to ask again, but where is the 1/4 coming from?

 

[SammyS]

so i should have
1/4\displaystyle \\\int_{0}^{4} \ \frac{4e^\sqrt{x}}{n\sqrt{x}}\ \

sorry to ask again, but where is the 1/4 coming from?

No.

The 4/n, which is Δx, becomes dx in for the integral version.

Added in Edit:

The 1/4 comes about because you need 4/n for Δx, so that you have dx to properly write tour integral.

Have you ever seen in your textbook, a case where an integral is written without a differential such as dx ?

 

[SammyS]

What you have is close to the right answer, but never quite there.

OK

Let's say we want to write a right Riemann sum to approximate the integral, \displaystyle \int_{0}^{4}f(x)\,dx \ .

Let's partition the interval [0, 4] into 100 equal-width intervals.

The width of each interval is Δx = 4/100 = 0.04 .

The height of each interval is f(x_k), where x_k is the right hand x value of the k-th interval.

We would then approximate the integral as:

\displaystyle f(0.04)\left(0.04\right)+f(0.08)\left(0.04\right)+f(0.12)\left(0.04\right)<br /> +\dots

\displaystyle =\sum_{k=1}^{100}f(x_k)\frac{4}{100}\ .<br />​

If you use n intervals instead, you have

\displaystyle \sum_{k=1}^{n}f(x_k)\frac{4}{n}\ .The function you're using already has the n in it, but not the 4 .

 

[HallsofIvy]

Homework Statement

The interval [0,4] is partitioned into n equal subintervals and ti, is chosen in [Xi-1, Xi] for each i.
What is:

Lim... inf
x->inf...Ʃ e^(sqrt ti)/N(sqrt ti)
...i=1

You cannot take the limit as "x goes to infinity"- there is NO "x" in your formula. I think you meant "n goes to infinity".

The Attempt at a Solution

It was problem we had to do in class as a group, teacher wasnt there to provide us an answer, and some people in my group have conflicting answers


What i did:

I wrote it as integral

4
∫ (e^(sqrt X))/(N(sqrt X))
0

The integral is the limit of the sums as n goes to infinity. There should be no "n" (or "N") in this.

I then use the substitution method:
U= sqrt X
du= 1/(2 sqrtX) dx
2du= 1/(sqrt X)

...4
2/n ∫ e^u = (2e^(sqrt 4))-(2e/n^(sqrt 0))
...0

= 2e^2-2
My answer = 2/n(e^2 - 1)



Can someone help me out, i don't really understand what we are suppose to do, and i apologize for the bad input. Appreciate it.

 

[smiggles]

What you have is close to the right answer, but never quite there.

OK

Let's say we want to write a right Riemann sum to approximate the integral, \displaystyle \int_{0}^{4}f(x)\,dx \ .

Let's partition the interval [0, 4] into 100 equal-width intervals.

The width of each interval is Δx = 4/100 = 0.04 .

The height of each interval is f(x_k), where x_k is the right hand x value of the k-th interval.

We would then approximate the integral as:

\displaystyle f(0.04)\left(0.04\right)+f(0.08)\left(0.04\right)+f(0.12)\left(0.04\right)<br /> +\dots

\displaystyle =\sum_{k=1}^{100}f(x_k)\frac{4}{100}\ .<br />​

If you use n intervals instead, you have

\displaystyle \sum_{k=1}^{n}f(x_k)\frac{4}{n}\ .


The function you're using already has the n in it, but not the 4 .

Thank you very much for your help, greatly appreciate it!

You cannot take the limit as "x goes to infinity"- there is NO "x" in your formula. I think you meant "n goes to infinity".


The integral is the limit of the sums as n goes to infinity. There should be no "n" (or "N") in this.

Thanks for clearing that up for me, apologize about the mistakes.

Glad there is still great people around to help others appreciate it!

 

[SammyS]

Thank you very much for your help, greatly appreciate it!

Thanks for clearing that up for me, apologize about the mistakes.

Glad there is still great people around to help others appreciate it!

So, what integral did you finally come up with ?

 

[smiggles]

The professor put up the answer, does it seem correct, how did the integral go from 0 to 4 TO 0 to 2. My original answer is very close to that, but i multiplied by 2 instead of multiplying by 1/2.

and how are you able to multiply 4/n by n/4, and what is the purpose to doing this?

 

[SammyS]

The professor put up the answer, does it seem correct, how did the integral go from 0 to 4 TO 0 to 2. My original answer is very close to that, but i multiplied by 2 instead of multiplying by 1/2.

and how are you able to multiply 4/n by n/4, and what is the purpose to doing this?

I agree with your professor's answer.

The limits of integration change due to doing integration by substitution.

Added in Edit:

So, why did you edit your post, erasing the image?


As for your question about 4/n and n/4:

Your professor took the expression \displaystyle \ \ \frac{e^\sqrt{t_i}}{n\sqrt{t_i}}\ \ and multiplied that by \displaystyle \ \ \frac{n}{4}\frac{4}{n}\ .\ Why can he do that ?

His motivation for doing that was that he needed \ \Delta x\ in his sum and \displaystyle \ \ \Delta x=\frac{4}{n}\ . He then kept the \ \Delta x\ isolated and combined the \displaystyle \ \ \frac{n}{4}\ \ with the given expression, giving:

\displaystyle \left(\frac{e^\sqrt{t_i}}{n\sqrt{t_i}}\frac{n}{4} \right)\frac{4}{n}=\frac{e^\sqrt{t_i}}{4\sqrt{t_i}}\Delta x​

You didn't have the 4 in the denominator.

(1/4)2 = 1/2 . So, there's a 1/2 rather than the 2 you had.