Show that $\int_{0}^{2}\sqrt{x^2+1}\,dx$ is at least $2\sqrt{2}$ without evaluating the integral

Question

Problem: Show that $\int_{0}^{2}\sqrt{x^2+1}\,dx$ is at least $2\sqrt{2}$ without evaluating the integral

By breaking the integral into two parts, and with some substitution of variables, I was able to obtain:

$\int_{0}^{2}\sqrt{x^2+1}\,dx = \int_{0}^{1}\sqrt{(x-1)^2+1}+\sqrt{(x+1)^2+1}\,dx$

Then I took the derivative of the integrand (on the right side of above equation) with respect to $x$ and was able to show through much algebra that it was greater than $0$ for $0<x<1$.

From this I concluded that the integrand is increasing over the interval $(0,1)$ and therefore the integrand is at least $2\sqrt{2}$ (i.e., the value of the integrand at $x=0$) over the interval $[0,1]$, and therefore:

$\int_{0}^{2}\sqrt{x^2+1}\,dx \geq 2\sqrt{2}$

This seems sound, but I wonder if there is some other observation to be made that would provide a simpler or more elegant solution.

Any ideas? Thanks!

Answer 1

The arc length of the graph of $\frac12t^2$ on $0\le t\le x$ is $$\int_0^x\sqrt{t^2+1}\,dt$$ and is longer than the line segment from $(0,0)$ to $(x,x^2/2)$. When $x=2$ this becomes $\int_0^2\sqrt{t^2+1}\,dt\ge\sqrt{2^2+2^2}=2\sqrt2$.

Answer 2

Since $y''=(x^2+1)^{-\frac{3}{2}}\geq 0$, the graph is concave-up.

The tangent line of the curve at $P(1,\sqrt{2})$ is $y=\frac{1}{\sqrt{2}}x+\frac{1}{\sqrt{2}}$ and it cuts the $y$-axis at $A(0,\frac{1}{\sqrt{2}})$ and the line $y=2$ at $B(2,\frac{3}{\sqrt{2}})$.

The area of the trapezoid $O(0,0), A(0,\frac{1}{\sqrt{2}}), B(2,\frac{3}{\sqrt{2}}), C(2,0)$, is equal to $\frac{\frac{1}{\sqrt{2}}+\frac{3}{\sqrt{2}}}{2}\times 2=2\sqrt{2}$.

Since graph is concave-up, the area under the curve is greater than the area of this trapezoid.

Answer 3

By the Cauchy-Schwarz inequality $$\int\limits_0^2\sqrt{x^2+1}\,dx\ge \int\limits_0^2{x\cdot 1+1\cdot 1\over \sqrt{2}}\,dx=2\sqrt{2}$$