What is the Limit of Summation Notation for a Given Interval?

[Biosyn]

Homework Statement

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Homework Equations

The Attempt at a Solution

I just plugged in ∞ for n

[2+$\frac{3}{∞}$]² ($\frac{3}{∞}$) =

[2+0]² (0) = 0Did I do the problem correctly? I might need a refresher on summation notations.

Here are the multiple choice answers:

a. 0
b. 1
c. 4
d. 39
e. 125

 

[4570562]

Seems right to me.

 

[Dick]

It's not right at all. You can't plug n=infinity into that. It looks like a Riemann sum approximation to an integral to me. None of the multiple choice answers that you've shown are correct either.

 

[Biosyn]

It's not right at all. You can't plug n=infinity into that. It looks like a Riemann sum approximation to an integral to me. None of the multiple choice answers that you've shown are correct either.

I couldn't fit all of the choices into the frame.

The choices:

a. 0
b. 1
c. 4
d. 39
e. 125

 

[Dick]

I couldn't fit all of the choices into the frame.

The choices:

a. 0
b. 1
c. 4
d. 39
e. 125

I'm not going to pick an answer for you. Show me how to get it. Put x=k/n and express that as a limiting sum for a Riemann integral over x.

 

[Biosyn]

Okay , I think I did it.


(3/n) = ΔX
ΔX = (b-a)/n

so, b=3 ; a=0

x_i = a + [i(b-a)]/n

x_i = [0 + (3k)/n + 2]²

f(x_i) = (x+2)²


The integral would be $^{3}_{0}$∫(x+2)²

 

[Ray Vickson]

Okay , I think I did it.


(3/n) = ΔX
ΔX = (b-a)/n

so, b=3 ; a=0

x_i = a + [i(b-a)]/n

x_i = [0 + (3k)/n + 2]²

f(x_i) = (x+2)²


The integral would be $^{3}_{0}$∫(x+2)²

One of the listed answers is correct.

RGV

 

[Dick]

Okay , I think I did it.


(3/n) = ΔX
ΔX = (b-a)/n

so, b=3 ; a=0

x_i = a + [i(b-a)]/n

x_i = [0 + (3k)/n + 2]²

f(x_i) = (x+2)²


The integral would be $^{3}_{0}$∫(x+2)²

Right.