1 Attachment(s)

Finite sums to approximate integrals

Attachment 22546

Hmm not sure if I copied that image properly. Anyways I'm really confused on what exactly I need to do. The question is related to an introduction to calculus course. Where you take an infinite number of sqaures to find the area under the curve. For example taking the right hand side of the subinterval, where you multiply 1/n and (1/n)^2 and keep adding until 1/n.(n/n)^2. which would solve to be (1/n)^3 [1^2 + 2^2 + 3^2 + ... + n^2].

However 1^2 + 2^2 + 3^2 + ... + n^2 = ( n(n+1) (2n+1) )/ 6 <-- not sure where this formula comes from

Then you multiply that by 1/n^3 then take the limit as n approaches infinity solve using l hospitals rule and you get 1/3 for the area for the right hand side

But the question i have to solve is 1^15 + 2^15. etc where do i find the formula for 1^15, or better yet, how do i work it out? and how does it relate to 1/n^15, 1/n^16, 1/n^17 from the question. Im so confused (Crying) Thanks so much for any tips you guys can see (Happy)

Re: Finite sums to approximate integrals

__Hint__

$\displaystyle \lim_{n\to \infty}\frac{1}{n}\sum_{k=1}^nf(k/n)=\int_0^1f(x)\;dx$.

Re: Finite sums to approximate integrals

hmm im not sure what that formula means, im only new to calculus, thanks anyways :D

Re: Finite sums to approximate integrals

Quote:

Originally Posted by

**insanic1** hmm im not sure what that formula means, im only new to calculus, thanks anyways :D

Well, it is not clear for me the exact context of this problem. The formula I provided you is a well known expression for the integral of a continuous function in the interval $\displaystyle [0,1]$ , which allows to compute easily some limits. For example, for your first limit:

$\displaystyle \displaystyle\begin{aligned}\displaystyle\lim_{n \to{+}\infty}\dfrac{1}{n^{16}}\left(1^{15}+2^{15}+ \ldots+n^{15}\right)&=\displaystyle\lim_{n \to{+}\infty}{}\dfrac{1}{n}\sum_{k=1}^n\left(\frac {k}{n}\right)^{15}\\&=\int_0^1x^{15}\;dx\\&=\left[\dfrac{x^{16}}{16}\right]_0^1\\&=\dfrac{1}{16}\end{aligned}$