Question

Prove that $4 \leq \int_{1}^{3} \sqrt{3+x^2} dx \leq 4\sqrt{3}$

Answer 1

The statement is true.

Answer 2

Let $f(x) = \sqrt{3+x^2}$. The function $f(x)$ is increasing on $[1, 3]$ because its derivative $f'(x) = \frac{x}{\sqrt{3+x^2}}$ is positive for $x \in [1, 3]$. The minimum value of $f(x)$ on $[1, 3]$ is $m = f(1) = \sqrt{3+1^2} = 2$. The maximum value of $f(x)$ on $[1, 3]$ is $M = f(3) = \sqrt{3+3^2} = \sqrt{12} = 2\sqrt{3}$. Using the property $m(b-a) \leq \int_{a}^{b} f(x) dx \leq M(b-a)$, with $a=1$ and $b=3$, we get: $2(3-1) \leq \int_{1}^{3} \sqrt{3+x^2} dx \leq 2\sqrt{3}(3-1)$ $4 \leq \int_{1}^{3} \sqrt{3+x^2} dx \leq 4\sqrt{3}$