Integral Line, I mean Limit!

$I_{ n }=\int _{ 0 }^{ 1 } \sqrt { \frac { 1 }{ x } +n^{ 2 }x^{ 2n } } dx\\ \sqrt { x } =t\\ \Rightarrow I_{ n }=2\int _{ 0 }^{ 1 } \sqrt { 1+n^{ 2 }t^{ 2(2n+1) } } dt\\ \Rightarrow I_{ n }=2\int _{ 0 }^{ 1 } \sqrt { 1+{ (nt^{ (2n+1) }) }^{ 2 } } dt$

Now, consider the function $nx^{ (2n+1) }$ . This function is the derivative of:

$f(x)=\frac { nx^{ (2n+2) } }{ 2n+2 }$

Hence our integral is of the form:

$I_{ n }=2\int _{ 0 }^{ 1 } \sqrt { 1+{ (\frac { df\left( t \right) }{ dt } ) }^{ 2 } } dt$

If we observe the above integral, we see that $I_n$ is simply twice the length of the curve $f(x)$ between $x=0$ and $x=1$ .

Now, as $n \to \infty \Rightarrow f(x)=\frac { x^{ (2n+2) } }{ 2 }$

If we draw the graph of this, it will look like this:

Where the blue line is $x=1$ , the green line is $y=1/2$ and the red line is the curve. Evidently, the total length of curve is $1+1/2=3/2$ .

Hence, $I_n=2(3/2)= \boxed{3}$

Split the interval of integral into two subintervals $[0,1]=[0,1-\epsilon]+[1-\epsilon,1] \ \ \ \ \ \ \ \epsilon \rightarrow 0^{+}$ For very large $n$ in the first interval we have $\frac{1}{x} \gg n^2 x^{2n}$ and in the second $n^2 x^{2n} \gg \frac{1}{x}$ .

The integral becomes $I=\int_{0}^{1-\epsilon} \frac{1}{\sqrt{x}} \text{d} x+\int_{1-\epsilon}^{1} n x^n \text{d}x$ $I=2\sqrt{1-\epsilon}+\frac{n}{n+1} (1-(1-\epsilon)^{n+1})$ Taking the limit of $I$ with $\epsilon \rightarrow 0^{+}$ and $n \rightarrow \infty$ gives $I=3$ .