Your method in general is good; let's pick up at step 4:
Using the summation formulas, we have:
and so:
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