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Question: https://gyazo.com/03dbf19bc8af2d526003aa04566a8cd1

If there exist $m , M \in R^+$ such that $m \leq f(x) \leq M \forall x \in [a,b]$, then $$m(b-a) \leq \int_{a}^{b} f(x)dx \leq M(b-a)$$ - The Integral Inequality property

We would just need to find m and M to prove this:

a = -1

b = 1

$m(2) = 2 = 1$

$M(2) = 2\sqrt{2} = \sqrt{2}$ So now we know that m = 1 and $M = \sqrt{2}$, then the property is true

Is my answer correct?

2 Answers 2

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You have figured out what $M$ and $m$ need to be to apply the inequality, but you have not shown that they obey $$m\le f(x) \le M$$ for all $x\in [a,b],$ which is a precondition for the inequality to hold.

So to complete the problem, you need to show that $1 \le \sqrt{1+x^2} \le \sqrt{2}$ for all $x\in [-1,1].$

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    How do I do the bottom part? , I mean I could simplify it: $$1^2 \leq 1 + x^2 \leq 2$$ $$0 \leq x^2 \leq 1$$ $$-1 \leq x \leq 1$$ ? Does that prove it? – 2017-01-24
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    No, not really, you're proving things backwards. You started with what you wanted ($1<(1+x^2) < 2$) and proved what you know (-1$y=\sqrt{1+x^2}$ on the interval $[-1,1]$. Where is it largest and smallest? What y values does it lie between? – 2017-01-24
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Yes, your answer is correct, since

$1 \le \sqrt{1+x^2} \le \sqrt{1+1} \le \sqrt{2}$

for $x \in [-1,1]$.