Expressing the limit of a sum as a definite integral

  • #1

Homework Statement


Express the following as a definite integral:

Express the attached limit as an integral.


The Attempt at a Solution


I have gotten as far as every part of the answer except the upper bound. the answer is:
10
(from 1 to 10) [x-4lnx]dx
1

since the definition of the definite integral is:
a
f(x)dx = lim Ʃ Δxif(x)
b________Δ→∞ i=1

i set Δxi = 9/n since that approaches zero. f(x) would be left to 1+9i/n - 4ln(1+9i/n)
so i set x = 1+9i/n.
since n approaches ∞ and the upper bound of the sum is ∞, i plugged ∞ in for i and n.
thats where I have trouble. ∞/∞ is undefined. when i plug 1 in i end up with 1 so that is the lower bound.

Homework Statement





Homework Equations





The Attempt at a Solution

 

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Answers and Replies

  • #2
34,475
10,597
I don't think your infinite sum converges for any i - you sum over i which grows like i^2 and the log-expression does not reduce this enough (just grows with i*log(i)).
If the sum is supposed to run from i=1 to n, this makes sense, and you get the maximal x-value simply by setting i=n.
 
  • #3
Dick
Science Advisor
Homework Helper
26,258
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Homework Statement


Express the following as a definite integral:

Express the attached limit as an integral.


The Attempt at a Solution


I have gotten as far as every part of the answer except the upper bound. the answer is:
10
(from 1 to 10) [x-4lnx]dx
1

since the definition of the definite integral is:
a
f(x)dx = lim Ʃ Δxif(x)
b________Δ→∞ i=1

i set Δxi = 9/n since that approaches zero. f(x) would be left to 1+9i/n - 4ln(1+9i/n)
so i set x = 1+9i/n.
since n approaches ∞ and the upper bound of the sum is ∞, i plugged ∞ in for i and n.
thats where I have trouble. ∞/∞ is undefined. when i plug 1 in i end up with 1 so that is the lower bound.

Homework Statement





Homework Equations





The Attempt at a Solution

There's a typo in the attached limit expression. The upper limit on the summation should be n. As written it doesn't approach anything. The sum by itself diverges.
 
Last edited:

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