Answer
$\int_{0}^{0.5} cos(x^2)~dx \gt \int_{0}^{0.5} cos(\sqrt{x})~dx$
Work Step by Step
On the interval $0 \leq x \leq 0.5$:
$0 \leq x^2 \leq \sqrt{x}$
$1 \geq cos(x^2) \geq cos(\sqrt{x})$
Therefore, by Property 7:
$\int_{0}^{0.5} cos(x^2)~dx \geq \int_{0}^{0.5} cos(\sqrt{x})~dx$
On the interval $0 \lt x \leq 0.5$:
$0 \lt x^2 \lt \sqrt{x}$
$1 \gt cos(x^2) \gt cos(\sqrt{x})$
Therefore:
$\int_{0}^{0.5} cos(x^2)~dx \gt \int_{0}^{0.5} cos(\sqrt{x})~dx$
