IIT JEE , AIEEE Forum - Mathematics , Algebra

konichiwa2x

very nice problem!

k..

(n=1000) Here is the proof.

let

consider this function in the interval

For every

, we have a rectangle with the interval

as its base and

as its height. The sum of the areas of these (upper) rectangles is certainly bigger than the area under the graph of

Now:
Sum of the areas of the rectangles =

, and
area under the graph of f(x) =

So we have

this time consider the rectangles with, again, the intervals

as their bases but this time

as their heights. this time the sum of the areas of these (lower) rectangles is certainly less than the area under the graph of

Now,
Sum of the areas of the rectangles =

Thus

So we've proved that

=