How to compute $\lim_{t \to \infty} \int_0^1 \sqrt{1+t^2 x^{2t-2}} dx$ without using arclength? In computing
$$\lim_{t \to \infty} \int_0^1 \sqrt{1+t^2 x^{2t-2}} dx,$$
realizing the integral is of the form $\int_0^1 \sqrt{1+f'(x)^2}dx$ with $f(x)=x^t$ allows one to use the triangle inequality to make estimates on the arclength of $f(x)=x^t$ as $t \to \infty$. From this one can conclude the limit is 2.
How can I do this without thinking of this as an arclength? What are other methods of evaluation?
 A: Note that we can write the integral of interest as
$$\begin{align}
\int_0^1 \sqrt{1+t^2x^{2t-2}}\,dx&=\int_0^1 tx^{t-1}\,dx+\int_{0}^{1} \left(\sqrt{1+t^2x^{2t-2}} -tx^{t-1} \right)\,dx\\\\
&=1+\int_{0}^1 \frac{1}{\sqrt{1+t^2x^{2t-1}}+tx^{t-1}}\,dx \tag1
\end{align}$$
It is trivial to see that $\left|\frac{1}{\sqrt{1+t^2x^{2t-1}}+tx^{t-1}}\right|\le 1$ for all $x\in [0,1]$ and $t\ge 0$.  Hence, the Dominated Convergence Theorem guarantees that 
$$\begin{align}
\lim_{t\to \infty}\int_0^1 \frac{1}{\sqrt{1+t^2x^{2t-1}}+tx^{t-1}}\,dx&=\int_0^1 \lim_{t\to \infty}\left(\frac{1}{\sqrt{1+t^2x^{2t-1}}+tx^{t-1}}\right)\,dx\\\\
&=1\tag 2
\end{align}$$

Finally, using the result from $(2)$ yields
$$\bbox[5px,border:2px solid #C0A000]{\lim_{t\to \infty}\int_0^1 \sqrt{1+t^2x^{2t-2}}\,dx=2}$$ 

And we're done!
A: The physicist's quick and dirty route:
Set $$q(t)=\int_0^1 dx \left(\sqrt{1+t^2x^{2t-2}}-1\right)$$
as $t\rightarrow \infty$ this integral is sharply peaked around $x=1$ and zero elsewhere. In this region the square root can be approximated by $tx^{t-1}$ by (the part of the integration range where $x\ll1$ yields a neglible contribution so we can just add it)
$$
q(t)\sim t\int_0^1 dx x^{t-1}=1
$$
since the original integral is $p(t)=\int_0^1dx+q(t)$

$$
\lim_{t\rightarrow\infty}p(t)=2
$$

A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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$Laplace\ Asymptotic\ Method:$

\begin{align}
&\color{#f00}{\lim_{t \to \infty}\int_{0}^{1}\root{1 + t^{2}x^{2t - 2}}\,\dd x} =
\lim_{t \to \infty}\int_{0}^{1}\root{1 + t^{2}\pars{1 - x}^{2t - 2}}\,\dd x
\\[5mm] = &\
\lim_{t \to \infty}\int_{0}^{1}
\exp\pars{{1 \over 2}\ln\pars{1 + t^{2}\bracks{1 - x}^{2t - 2}}}\,\dd x
\\[5mm] = &\
\lim_{t \to \infty}\int_{0}^{\infty}
\exp\pars{{1 \over 2}\ln\pars{1 + t^{2}} -
{t^{3} - t^{2} \over 2\pars{1 + t^{2}}}\, x}\,\dd x =
\lim_{t \to \infty}\bracks{%
\root{1 + t^{2}}\,{2\pars{1 + t^{2}} \over \pars{t - 1}t^{2}}} = \color{#f00}{2}
\end{align}
A: I will provide a not-totally-rigorous direct proof. I'd love the see the fully rigorous details.
Recall Gauss' hypergeometric function
$${}_2F_1 (a,b;c;z) = \displaystyle\sum_{k=0}^{\infty} \dfrac{a^{\overline{k}}b^{\overline{k}}}{c^{\overline{k}}} \dfrac{z^k}{k!},$$
where $\xi^{\overline{k}}$ denotes the rising factorial
$$\xi^{\overline{k}}=\dfrac{\Gamma(\xi+k)}{\Gamma(\xi)}=\xi (\xi+1) \ldots (\xi+k-1).$$
In particular note that we take $\xi^{\overline{0}}$ to be $1$. This shows that 
$${}_2F_1(a,0;c;z) = 1,$$
a constant function. For our purposes, 
$$\displaystyle\lim_{|z| \rightarrow \infty} {}_2F_1(a,0;c;z)=1.$$
Wolfram alpha reports the necessary antiderivative is
$$\displaystyle\int \sqrt{1 + ax^b} \mathrm{d}x = \dfrac{xb {}_2F_1\left( \frac{1}{2}, \frac{1}{b};1+\frac{1}{b};-ax^b \right)}{b+2}+\dfrac{2x\sqrt{ax^b+1}}{b+2}+C.$$
So with $a=t^2$ and $b=2t-2$, the fundamental theorem of calculus says
$$\begin{array}{ll}
\displaystyle\lim_{t\to \infty}\int_0^1 \sqrt{1+t^2x^{2t-2}} \mathrm{d}x &= \displaystyle\lim_{t \rightarrow \infty} \left[\dfrac{(2t-2){}_2F_1\left(\frac{1}{2},\frac{1}{2t-2};1+\frac{1}{2t-2};-t^2\right)}{2t} + \dfrac{2\sqrt{t^2+1}}{2t}{}\right] \\
&=\displaystyle\lim_{t \rightarrow \infty}  \left[ \left(\dfrac{t-1}{t} \right) {}_2F_1\left( \dfrac{1}{2}, \dfrac{1}{2t-2};1+\dfrac{1}{2t-2};-t^2 \right) + \dfrac{\sqrt{t^2+1}}{t}\right].
\end{array}$$
Since
$$\displaystyle\lim_{t \rightarrow \infty} \dfrac{t-1}{t} = \displaystyle\lim_{t \rightarrow \infty}\dfrac{\sqrt{t^2+1}}{t}=1,$$
to finally answer the question, we must discuss the limiting behavior of the ${}_2F_1$ function.
A rigorous approach would consider asymptotic formulas of the hypergeometric series and deduce the limit from them. We will relax rigor and just take $t \rightarrow \infty$ in all parameters to ${}_2F_1$ suggesting limiting value should be
$${}_2F_1 \left( \dfrac{1}{2}, 0; 1; -\infty \right)=1.$$
This means we would have
$$\displaystyle\lim_{t \rightarrow \infty} \displaystyle\int_0^1 \sqrt{1+t^2 x^2} \mathrm{d}x = 1+1=2,$$
as desired.
