Comparing Areas under Curves I remembered back in high school AP Calculus class, we're taught that for a series:
$$\int^\infty_1\frac{1}{x^n}dx:n\in\mathbb{R}_{\geq2}\implies\text{The integral converges.}$$
Now, let's compare $$\int^\infty_1\frac{1}{x^2}dx\text{ and }\int^\infty_1\frac{1}{x^3}dx\text{.}$$
Of course, the second integral converges "faster" since it's cubed, and the area under the curve would be smaller in value than the first integral.
This is what's bothering me:  I found out that 
$$\int^\infty_{1/2}\frac{1}{x^2}dx=\int^\infty_{1/2}\frac{1}{x^3}dx<\int^\infty_{1/2}\frac{1}{x^4}dx$$
Can someone explain to me when is this happening, and how can one prove that the fact this is right?
Thanks!
 A: The difference is that in your second set of integrals, the lower limit is less than one. In this range, $1/x^2 < 1/x^3 < 1/x^4$, and has an effect on your overall integral. Try breaking your second set of integrals into two pieces each: Over $[1/2, 1]$, then over $[1, \infty]$ and see what happens.
When the exponent $p$ and $q$ are both at least one, and $p \lt q$ then when $x \ge 1$ you get $x^p \le x^q$ so $\displaystyle \frac{1}{x^p} \ge \frac{1}{x^q}$. On the other hand when $0 \lt x \lt 1$, then $x^p \ge x^q$ so $\displaystyle \frac{1}{x^p} \le \frac{1}{x^q}$.
(Of course everything is equal when $x = 1$).
You can see how this works if you integrate over an interval containing $x = 1$, for example split out
$$\int \limits _a^b \frac{1}{x^p}\text{ }dx = \int \limits _a^1 \frac{1}{x^p}\text{ }dx + \int \limits _1^b \frac{1}{x^p}\text{ }dx$$
Vary $p$ and look at what happens to each piece individually...
So for your examples, you can examine the four integrals
$\displaystyle \int \limits _{1/2}^1 \frac{1}{x^2}\text{ }dx$,
$\displaystyle \int \limits _1^\infty \frac{1}{x^2}\text{ }dx$,
$\displaystyle \int \limits _{1/2}^1 \frac{1}{x^3}\text{ }dx$, and
$\displaystyle \int \limits _1^\infty \frac{1}{x^3}\text{ }dx$
A: $$\text{The main reason that: }\int^\infty_{1/2}\frac{1}{x^2}dx=\int^\infty_{1/2}\frac{1}{x^3}dx$$
$$\text{is because although that }\int^\infty_1\frac{1}{x^2}dx>\int^\infty_1\frac{1}{x^3}dx$$
$$\text{remember that }\int^1_{1/2}\frac{1}{x^2}dx<\int^1_{1/2}\frac{1}{x^3}dx\text{.}$$
$$\text{So, in this case: }\int^\infty_1\frac{1}{x^2}dx+\int^1_{1/2}\frac{1}{x^2}dx=\int^\infty_1\frac{1}{x^3}dx+\int^1_{1/2}\frac{1}{x^3}dx,$$
$$\text{which means that: }\int^\infty_{1/2}\frac{1}{x^2}dx=\int^\infty_{1/2}\frac{1}{x^3}dx$$
