Write the definite integral for the summation:
n
Lim Σ (1+ k/n)^2 * (1/n)
n->∞ k=1
1
A. ∫ x^2 dx
0
2
B. ∫ (x+1)^2 dx
1
2
C. ∫ x^2 dx
1
Thank you in advance for the help!
1
D. ∫ x^2 dx
2
I'm sorry if the problem looks a bit confusing. It was a little hard to type it.
The upper bound of sigma notation is n and the lower bound is k=1.
*And the upper and lower bound of choices A through D are the numbers that I accidentally placed right above and below each Choice letter.
The problem of typing something like an integral is that this format
does not accept spaces, so we have to work around that
e.g.
B could be typed like this:
∫ (x+1)^2 dx , x goes from 1 to 2
Wolfram accepts that terminology.
www.wolframalpha.com/input/?i=%E2%88%AB+(x%2B1)%5E2+dx+,+x+goes+from+1+to+2
Ok, I will do that one
∫ (x+1)^2 dx , x goes from 1 to 2
= (1/3)(x+1)^3 | from 1 to 2
= (1/3)(3^3) - (1/3)2^3
= 27/3 - 8/3 = 19/3
for the others, you should be able to integrate polynomials of this type
Since the typing is most likely very confusing, here is the link with the pictures of the problems. 👍
docs.google.com/document/d/1ANn3kbGwHEal4-dyBMSt0q1t6bXKUZqbf6zEPxT48CA/edit?usp=sharing
Make sure to add the, h t t p s : / / before the link. I guess Jiskha doesn't allow links to be posted.
@Reiny
So I have to integrate each of them? Or are you saying that B is the right option(or did you use that as an example/)
You know that the area under a curve (the integral) can be approximated by a Riemann sum of rectangles, right?
Now consider this sum for for some value of n. You have
Σ (1+ k/n)^2 * (1/n)
The height of each rectangle is 1 + k/n
the width is 1/n
So, you have divided an interval of width 1 into n strips.
You are using the right-endpoint of each subinterval (since k goes from 1..n)
Consider the case where n=3. Your sum is
[(1+1/3)^2 + (1+2/3)^2 + (1+3/3)^2 ] (1/3) ≈ ∫[1,2] x^2 dx
To find the definite integral for the given summation, we can convert it into a Riemann sum first.
In the summation, we have a sum of terms (1 + k/n)^2 * (1/n) where the range of k is from 1 to n.
First, let's rewrite the given sum in the form of a Riemann sum:
∑ (1 + k/n)^2 * (1/n)
To convert it into a definite integral, we can rewrite the terms (1 + k/n)^2 * (1/n) as a function of x, where x is k/n:
(1 + k/n)^2 * (1/n) = (1 + x)^2 * (1/n)
Now, we need to determine the limits of the integral based on the given limit of the summation:
n -> ∞
As n approaches infinity, x = k/n approaches zero. So, the lower limit of the integral will be 0.
For the upper limit of the integral, as n approaches infinity, x = k/n also approaches 1. So, the upper limit of the integral will be 1.
Now, we convert the summation into a definite integral:
Lim(n→∞) ∑ (1 + k/n)^2 * (1/n)
k=1
= Lim(n→∞) ∫ (1 + x)^2 * (1/n) dx
0 to 1
= ∫ (1 + x)^2 dx
0 to 1
Let's simplify the integral:
= ∫ (1 + 2x + x^2) dx
0 to 1
= [x + x^2/2 + x^3/3] from 0 to 1
= (1 + 1/2 + 1/3) - (0 + 0 + 0)
= 11/6
So, the definite integral for the given summation is:
∫ (1 + x)^2 dx from 0 to 1, which is equal to 11/6.
Therefore, the correct answer is:
2
C. ∫ x^2 dx
1