Question

Find a function f for which the limit below represents the area under f on the interval [0, a].
Here we assume the rectangles used have equal widths and right endpoints as sample points.

http://i49.tinypic.com/20ua7i9.jpg

Answer 1

any guesses for this one?

Answer 2

no guesses? i guess \(f(x)=x^2\)

Answer 3

on the interval \([0,1]\) divide the interval in to \(n\) parts each of length \(\frac{1}{n}\)

Answer 4

Just x^2?
I thought it had to be more than that though

Answer 5

the first simple point will be \(\frac{1}{n}\) the second will be \(\frac{2}{n}\) and the \(i\)th will be \(\frac{i}{n}\)

Answer 6

then for each \(i\), \(f(\frac{i}{n})=\frac{i^2}{n^2}\)

Answer 7

oh wow! Dang. I hate this notation. It makes it ten times more complicated than it needs to be.

Answer 8

multiply each by \(\frac{1}{n}\) and add the up you get
\[\sum\frac{i^2}{n^3}\]

Answer 9

well don't be hoodwinked in to thinking these are all that easy
usually it is much harder to tell

Answer 10

I have another one a lot like this. with a sqaure root. but I'm not sure about it

Answer 11

but the first clue should have been the \(i^2\)
on the other hand if the interval had be \([1,2]\) it would have been completely different

Answer 12

interesting, because the i and n values would be different right?

Answer 13

because a-b / n ?