I wanted to ask a question about an equation, $A(n)$, where $n \in \mathbb{N}$ be the area in the first quadrant bound by $y = x$ and $y = x^n$. Would the following be true:
$A(n)=A(\frac{1}{n})$
- $\displaystyle \lim_{n \to \infty} = 1$?
- $A(n)=A(\frac{1}{n})$?
In general I visualize $y=x$ and $y=x^n$ as taking up most of the first quadrant, but takes up less space as the $x$ value increases. Does this come into play in any way?
It is possible to answer both of these questions without actually working out the integrals, albeit unrigorously.
For the first, think of the shape of the curve $y=x^n$ as $n$ increases. You'll notice the curve becomes more "square". Flatter close to $x=0$ then steeper as $x$ gets close to $1$ and beyond. Taken to the limit, the bounded area will tend toward forming an isosceles right triangle with both catheti (perpendicular sides) equal to $1$ and a hypotenuse formed by a segment of the line of identity $y=x$ with length $\sqrt 2$. That has area $\frac 12$, and it is the same value that you'd calculate using calculus (you can refer to the formula in Kavi's comment and take the limit quite easily).
For the second, note that if $f(x) = x^n$, then $f^{-1}(x) = x^{\frac 1n}$. The graph of the inverse function is found by reflecting the graph of the original function about the line of identity, and the line of identity itself remains invariant. Which means the area does not change when you replace $n$ with $\frac 1n$. You will, of course, arrive at the same result using the precise formula you get after integration and imposing bounds (again, refer to Kavi's formula and replace $n$ by its reciprocal).
So, first one false, second one true.