Answered: "Use the properties of integrals to verify the..."

High School
Algebra
Pre-Algebra
Equations, expressions, and inequalitie

rastafarral6

Answered question

2021-12-02

Use the properties of integrals to verify the inequality without evaluating the integrals.

2 \leq \int_{- 1}^{1} \sqrt{1 + x^{2}} d x \leq 2 \sqrt{2}

Answer & Explanation

Clara Clark

Beginner2021-12-03Added 17 answers

{For $- 1 \leq x \leq 1$ , we can say
$0 \leq x^{2} \leq 1$
$1 \leq 1 + x^{2} \leq 2$
$\sqrt{1} \leq \sqrt{1 + x^{2}} \leq \sqrt{2}$ }
The Comparison Property of Integrals states that we have:
$1 \cdot (1 - (- 1)) \leq \int_{- 1}^{1} \sqrt{1 + x^{2}} d x \leq \sqrt{2} \cdot (1 - (- 1))$
$2 \leq \int_{- 1}^{1} \sqrt{1 + x^{2}} d x \leq 2 \sqrt{2}$
Therefore proven
We made use of property no.
If $m \leq f (x) \leq M f or a \leq x \leq b$ , then
$m (b - a) \leq \int_{a}^{b} f (x) d x \leq M (b - a)$
Result:
Hint: $\sqrt{1} \leq \sqrt{1 + x^{2}} \leq \sqrt{2}$

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