## Calculus: Early Transcendentals 8th Edition

$2 \leq \int_{-1}^{1} \sqrt{1+x^2}~dx \leq 2\sqrt{2}$
Note the following: $\int_{-1}^{1} \sqrt{1+x^2}~dx = \int_{-1}^{0} \sqrt{1+x^2}~dx+\int_{0}^{1} \sqrt{1+x^2}~dx = 2\int_{0}^{1} \sqrt{1+x^2}~dx$ On the interval $0 \leq x \leq 1$: $0 \leq x^2 \leq 1$ $1 \leq 1+x^2 \leq 2$ $\sqrt{1} \leq \sqrt{1+x^2} \leq \sqrt{2}$ $1 \leq \sqrt{1+x^2} \leq \sqrt{2}$ Therefore, by the Comparison Property 8: $1 \leq \int_{0}^{1} \sqrt{1+x^2}~dx \leq \sqrt{2}$ $2 \leq 2\int_{0}^{1} \sqrt{1+x^2}~dx \leq 2\sqrt{2}$ $2 \leq \int_{-1}^{1} \sqrt{1+x^2}~dx \leq 2\sqrt{2}$