# Need help with a simple integral proof

1. Jul 23, 2013

### tolove

f is a positive decreasing function of x, f(n) = an for all n:

$\int\stackrel{n+1}{1} f(x) dx$ $\leq$ a$_{1}$ + a$_{2}$ + $\cdots$ + a$_{n}$

Why is this true? I think I'm missing a fundamental understanding of summation, since the reasoning for this step was left out in my book.

[edited out some nonsense]

Thanks for any help!

Last edited: Jul 23, 2013
2. Jul 23, 2013

### hilbert2

Write the left side of the inequality as $\int^{2}_{1}f(x)dx+\int^{3}_{2}f(x)dx+\dots\int^{n+1}_{n}f(x)dx$ and the right side as $\int^{2}_{1}a_{1}dx+\int^{3}_{2}a_{2}dx+\dots\int^{n+1}_{n}a_{n}dx$ and show that each term on the LHS is smaller than the corresponding term on the RHS. Note that $\int^{b}_{a}f(x)dx\leq\int^{b}_{a}g(x)dx$ if $f(x)\leq g(x)$ on interval [a,b].

3. Jul 23, 2013

### tolove

Alright. I've got this, I think.

In order to make sense of the series, I should represent it as a piecewise graph with height an and a width Δx, where Δx is always 1. The area of each piece of such a graph will be an*Δx, which is just an.

If Ʃ an is continuous, etc, then I can compare it to a function, f. f(x) is always going to dip underneath (in the case of decreasing functions) or be equal to the horizontal line y = an for each interval. Thus the area of f(x) is always less than the area of the piecewise graph. And, when calculating the area of f(x), I need to go out one more step, n+1, since the Δx from the piecewise graph steps out one further.

Thank you very much!