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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?

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    This follows by the obvious inequality $$\min_{x \in [a,b]} f(x) \le f(x) \le \max_{x \in [a,b]} f(x)$$ which can be integrated in $x \in [a,b]$ to get $$(b-a) \min_{x \in [a,b]} f(x) \le \int_a^b f(x) dx \le (b-a)\max_{x \in [a,b]}f(x)$$ – 2017-01-15
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    How exactly does this have anything to do with the title? – 2017-01-15
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    Sorry ... looks like i left an old question unfinished, haha ill fix – 2017-01-15

2 Answers 2

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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

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

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    How did you choose your values for $x$ to be $1$ and $0$? – 2017-01-15
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    @KSplitX Extrema occur when the first derivative is $0$ or at the end points. Sorry for the late reply. – 2017-01-16
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    So the valeus that you plug in, must they be extrema? – 2017-01-17
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    Yes, since it might be min or max? – 2017-01-17
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    @KSplitX Yes, that is an extrema by definition. – 2017-01-17
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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.

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    Thanks, but I dont think I learned that yet – 2017-01-15
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    I put my solution exactly for popularisation of the Hermite-Hadamard inequality. Please see it on Wikipedia. You need not prove it. Just learn there is something like that, maybe it will be useful in the future. – 2017-01-15