Question

http://i48.tinypic.com/2ljot3n.jpg

Answer 1

ok now i am experienced enough to ask what you did, so we can check why it didn't work

Answer 2

or we could ignore the directions and just add

Answer 3

try
\[\int_0^1(8+16x^2)dx\]

Answer 4

or else add this mess up, it is not that hard

Answer 5

for example
\[\frac{8}{n}\sum_{i=1}^n 1=8\] and
\[\frac{16}{n^3}\sum i^2=\frac{16}{n^3}\frac{n(n+1)(2n+1)}{6}\]

Answer 6

limit of the second one is \(\frac{32}{6}=\frac{16}{3}\) add to 8 and get \(\frac{40}{3}\)

Answer 7

Oh jeez. I didn't add the 8. XP thank you!

Answer 8

Sorry I didn't reply earlier