This probably as nearly a stupid question as the one that isn't asked, but I've always thought the area under the curve is equal to the integral. So Given the function $\dfrac{1}{x^2}$ which is symetric about $y = x$, why isn't the integral from $1$ to $\infty$ of $\dfrac{1}{x^2}$ equal to the $\displaystyle\int_0^1\dfrac{1}{x^2}\,dx$ minus $1$ The unit square under integrated area. \begin{align} \int_1^\infty\dfrac{1}{x^2}\,dx &= 1 - 0 = 1 & \text{and}&& \int_0^1\dfrac{1}{x^2}\,dx &= \infty - 1 =\infty \end{align} At least when using the Reimann improper integral. I also tried Lebesgue and got the same answers.
Any help would be greatly appreciated. Thanks for taking the time to read my question.
There is an isometry between the domain $D$ bounded by the curve $y=1/x$ for $x$ in $(0,1]$, the segment $[0,1]\times\{1\}$ and the halfline $\{0\}\times[1,+\infty)$, and the domain $E$ bounded by the curve $y=1/x$ for $x$ in $[1,+\infty)$, the segment $\{1\}\times[0,1]$ and the halfline $[1,+\infty)\times\{0\}$.
But the area of $E$ is $\int\limits_1^\infty\frac1{x}\mathrm dx=+\infty$. On the other hand, consider the square $K=[0,1]\times[0,1]$, then $K$ and $D$ are disjoint and the area of $K\cup D$ is $\int\limits_0^1\frac1{x}\mathrm dx=+\infty$.
Thus, if one insists on translating the isometry between $D$ and $E$ in terms of areas, one should write something like $+\infty=+\infty-1$... which, despite some uneasy feelings the RHS may bring to the reader, happens to be true.