anonymous
  • anonymous
Write the definite integral for the summation: the limit as n goes to infinity of the summation from k equals 1 to n of the product of the square of the quantity 1 plus k over n squared and 1 over n.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
https://gyazo.com/478f27e7e70fd5a27beb5d16043fe87e
anonymous
  • anonymous
As n approaches to infinity the values get closer to 0
Miracrown
  • Miracrown
Indeed, As n approaches to infinity the values get closer to 0

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Sorry went to use the restroom.
anonymous
  • anonymous
And now that I take a good look what I was thinking would make no sense here.
jim_thompson5910
  • jim_thompson5910
Hint: \[\Large \lim_{n \to \infty}\sum_{k = 1}^{n}\left(1+\frac{k}{n}\right)^2\left(\frac{1}{n}\right)\] \[\Large \lim_{n \to \infty}\sum_{k = 1}^{n}\left(1+\frac{1}{n}*k\right)^2\left(\frac{1}{n}\right)\] \[\Large \lim_{n \to \infty}\sum_{k = 1}^{n}\left({\color{red}{1}}+{\color{blue}{\frac{1}{n}}}*k\right)^2\left({\color{blue}{\frac{1}{n}}}\right)\] \[\Large \lim_{n \to \infty}\sum_{k = 1}^{n}\left({\color{red}{a}}+{\color{blue}{\Delta x}}*k\right)^2\left({\color{blue}{\Delta x}}\right)\]
jim_thompson5910
  • jim_thompson5910
Keep in mind that \[\Large x_k = a+\Delta x*k\] \[\Large \Delta x = \frac{b-a}{n}\]
anonymous
  • anonymous
So b=2
jim_thompson5910
  • jim_thompson5910
yep
anonymous
  • anonymous
But what is Xk for?
jim_thompson5910
  • jim_thompson5910
xk represents a general term in the sequence x1,x2,x3, ..., xn
jim_thompson5910
  • jim_thompson5910
the idea is that you can find the approx area under the curve by breaking up the interval into n equal pieces and finding the area of each rectangular piece
jim_thompson5910
  • jim_thompson5910
this might help visualize an example https://www.math.hmc.edu/calculus/tutorials/riemann_sums/gif/figure6.gif
anonymous
  • anonymous
But Xk would be equal to what's in the first parenthesis so how will that help me when writing the integral?
anonymous
  • anonymous
Integral is equal to area under curve
jim_thompson5910
  • jim_thompson5910
Hint: f(x) ----> f(xk)
anonymous
  • anonymous
What?
jim_thompson5910
  • jim_thompson5910
Do you see how I got\[\Large \lim_{n \to \infty}\sum_{k = 1}^{n}\left({\color{red}{1}}+{\color{blue}{\frac{1}{n}}}*k\right)^2\left({\color{blue}{\frac{1}{n}}}\right)\]
anonymous
  • anonymous
Yes I understand the concept just not how it will be applied
jim_thompson5910
  • jim_thompson5910
that will turn into \[\Large \lim_{n \to \infty}\sum_{k = 1}^{n}\left(x_k\right)^2\left({\color{black}{\frac{1}{n}}}\right)\]
jim_thompson5910
  • jim_thompson5910
the \(\LARGE (x_k)^2\) portion represents the f(x) function the remaining bit is the delta x
anonymous
  • anonymous
Ohhhhhhhh I feel stupid now :/
anonymous
  • anonymous
Thanks
Miracrown
  • Miracrown
Don't feel stupid. You're still learning :)
jim_thompson5910
  • jim_thompson5910
we have n rectangles each rectangle is delta x = 1/n units wide the height of each rectangle is equal to f(x_k) = (1+k/n)^2 units
anonymous
  • anonymous
Got it Thanks

Looking for something else?

Not the answer you are looking for? Search for more explanations.