Prove that area under curve by rectangle is less than integration

[basil]

hi all,

I am suppose to compare the area of curve y=x2 with rectangles beneath that curve to show that,

1/2 + 1/3 + ...+ 1/n < log(n+1)

i believe this some sort of harmonic series. Is there a way around this problem?

Regards

[quasar987]

I am suppose to compare the area of curve y=x2 [...]

Don't you mean y=1/x?

[feynman137]

I think it is y=1/x. So, take an arbitrary interval [0,l] and employ riemann sums: divide it in n subintervals, each delta x= l/n wide. The area of all rectangles of basis delta x and height 1/i(delta x), where i runs from 1 to n, is an approximation downwards of the area under the curve. In particular, it will be less than the area from x=delta x to x=l, which is just log n, plus the area of the first rectangle, that is 1. So, sum(from i=1 to n) 1/i <log n +1 and we are done.

[lavinia]

hi all,

I am suppose to compare the area of curve y=x2 with rectangles beneath that curve to show that,

1/2 + 1/3 + ...+ 1/n < log(n+1)

i believe this some sort of harmonic series. Is there a way around this problem?

Regards

1/2 + 1/3 + ...+ 1/n < or = log(n+1) is by definition of lower sum. To prove strict inequality subdivide more and show that the sum is bigger.

[basil]

Oh yes, its

\frac{1}{x}

Nevertheless, the explanation was beneficial. Thanks guys.