Show $2 \leq \int^1_{-1}\sqrt{1+x^2}dx \leq 2\sqrt{2}$ without evaluating

Question

Use the properties of definite integrals to verify that:

$$2 \leq \int^1_{-1}\sqrt{1+x^2}dx \leq 2\sqrt{2}$$

without evaluating.

I have no idea how to begin. I guess I could draw a graph, but thats kind of same as verification. Any hints would be nice?

Answer 1

Notice that

$$\int_{-1}^1\sqrt{1+x^2}\ dx<\int_{-1}^1\sqrt{1+1^2}\ dx=2\sqrt2$$

Likewise,

$$\int_{-1}^1\sqrt{1+x^2}\ dx>\int_{-1}^1\sqrt{1+0^2}\ dx=2$$

where we used

$$\int_a^b\min_{t\in(a,b)}f(t)\ dx\le\int_a^bf(x)\ dx\le\int_a^b\max_{t\in(a,b)}f(t)\ dt$$

Answer 2

Another solution could be given by the Hermite-Hadamard inequality.

It is easy to verify that $f(x)=\sqrt{\strut 1+x^2}$ is convex (by $f''>0$). Then $$f(0)\leqslant \frac{1}{2}\int\limits_{-1}^1 f(x)\text{d}x\leqslant \frac{f(-1)+f(1)}{2}.$$ For our function we have $$1\leqslant \frac{1}{2}\int\limits_{-1}^1\sqrt{\strut 1+x^2}\text{d}x\leqslant \sqrt{2},$$ which is the desired inequality.