Checking over limit process

Evaluate the definite integral by the limit process:

_{1}∫^{2}(x^{2}+1)dx

Answer from book: 10/3

This is what I did order:

1. I used the right justified Riemann Sum (by the way, on the top of ∑ is n, and the bottom is i=1, so like _{i=1}∑^{n}, except the n is on the top and the i=1 is on the bottom, but I don't know how to type it like that, so just letting you know that I'll do ∑ alone)

(2 - 1)/n ∑ f(1 + ((2-1)/n)i)

2. I simplified it

1/n ∑ f(1 + (1/n)i)

3. I replaced f with the original equation

1/n ∑ (1 + (1/n)i)^{2 }+ 1

4. I foiled it

1/n ∑ (1 + (2/n)i + (1/n^{2})i^{2}) + 1

5. I used the sigma notation and multiplied the 'n' into the function

1/n [(1n + 2n(n(n+1))/2) + 1/n^{2} ((n(n+1)(2n+1))/6) + 1n]

6. I multiplied the 1/n outside into everything inside

1 + 2/n^{2} (n^{2}/2 + n/2) + 1/n^{3} (2n^{3}/6 + 3n^{3}/6 + n/6) + 1

7. I got rid of everything that had an n inside it, since the limit was n -> infinity, so the closer to 0

That meant I was left with:

1 + 2 + 2/6 + 1, which equals 6/6 + 12/6 + 2/6 + 6/6, which equals 26/6, which equals **13/3**

I think I might have done something wrong, can someone explain to me if I was doing this process correctly, but there might have been an error as well, so if you can help me point that out where I went wrong.

Limit process has so many multiplications and a long process, so I may have gotten lost in some ways.

Thanks

Re: Checking over limit process

Your method in general is good; let's pick up at step 4:

$\displaystyle S_n=\frac{1}{n}\sum_{k=1}^n\left(2+\frac{2k}{n}+ \frac{k^2}{n^2} \right)$

Using the summation formulas, we have:

$\displaystyle S_n=\frac{1}{n}\left(2n+\frac{2}{n}\cdot\frac{n(n+ 1)}{2}+\frac{1}{n^2}\cdot\frac{n(n+1)(2n+1)}{6} \right)$

$\displaystyle S_n=2+\frac{(n+1)}{n}+\frac{(n+1)(2n+1)}{6n^2}$

and so:

$\displaystyle \lim_{n\to\infty}S_n=2+1+\frac{1}{3}=\frac{10}{3}$

Re: Checking over limit process

Quote:

Originally Posted by

**MarkFL2** Your method in general is good; let's pick up at step 4:

$\displaystyle S_n=\frac{1}{n}\sum_{k=1}^n\left(2+\frac{2k}{n}+ \frac{k^2}{n^2} \right)$

Using the summation formulas, we have:

$\displaystyle S_n=\frac{1}{n}\left(2n+\frac{2}{n}\cdot\frac{n(n+ 1)}{2}+\frac{1}{n^2}\cdot\frac{n(n+1)(2n+1)}{6} \right)$

$\displaystyle S_n=2+\frac{(n+1)}{n}+\frac{(n+1)(2n+1)}{6n^2}$

and so:

$\displaystyle \lim_{n\to\infty}S_n=2+1+\frac{1}{3}=\frac{10}{3}$

Ooo, I see thanks!

I might have messed up with those n(n+1)/2 thing, since I turned it into n^2/2 + n/2, so I might have messed up the multiplications.

But after seeing it in a shorter form, I can see how you got to 10/3

Thanks!

Re: Checking over limit process

Quote:

Originally Posted by

**MarkFL2** Your method in general is good; let's pick up at step 4:

$\displaystyle S_n=\frac{1}{n}\sum_{k=1}^n\left(2+\frac{2k}{n}+ \frac{k^2}{n^2} \right)$

Using the summation formulas, we have:

$\displaystyle S_n=\frac{1}{n}\left(2n+\frac{2}{n}\cdot\frac{n(n+ 1)}{2}+\frac{1}{n^2}\cdot\frac{n(n+1)(2n+1)}{6} \right)$

$\displaystyle S_n=2+\frac{(n+1)}{n}+\frac{(n+1)(2n+1)}{6n^2}$

and so:

$\displaystyle \lim_{n\to\infty}S_n=2+1+\frac{1}{3}=\frac{10}{3}$

Just wondering real quick, how did you get 2n on the left side of the equation.

I'm a bit confused, like after you foiled it, the left ended up a 1? o.o

Re: Checking over limit process

$\displaystyle \left(1+\frac{1}{n}k \right)^2+1=1+\frac{2}{n}k+\frac{k^2}{n^2}+1=2+ \frac{2}{n}k+ \frac{k^2}{n^2}$

Re: Checking over limit process

Quote:

Originally Posted by

**MarkFL2** $\displaystyle \left(1+\frac{1}{n}k \right)^2+1=1+\frac{2}{n}k+\frac{k^2}{n^2}+1=2+ \frac{2}{n}k+ \frac{k^2}{n^2}$

Oh right!

I forgot about the 1 at the end!

Thanks!