Definite integral properties to solve problems without evaluating

Question

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?

Answer 1

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

Answer 2

Yes, your answer is correct, since

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

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