Answer
$\int_{0}^{1} \sqrt{1+x^2}~dx \leq \int_{0}^{1} \sqrt{1+x}~dx$
Work Step by Step
On the interval $0 \leq x \leq 1$:
$x^2 \leq x$
$1+x^2 \leq 1+x$
$\sqrt{1+x^2} \leq \sqrt{1+x}$
Therefore, by the comparison property:
$\int_{0}^{1} \sqrt{1+x^2}~dx \leq \int_{0}^{1} \sqrt{1+x}~dx$
