Answer
(a) $cos(x^2) \geq cos~x$
(b) $\int_{0}^{\pi/6}cos(x^2)~dx \geq \frac{1}{2}$
Work Step by Step
(a) On the interval $~~0 \leq x \leq 1$:
$0 \leq x^2 \leq x$
$1 \geq cos(x^2) \geq cos~x$
(b) Note that $\frac{\pi}{6} \lt 1$
Therefore:
$\int_{0}^{\pi/6}cos(x^2)~dx \geq \int_{0}^{\pi/6}cos~x~dx$
$\int_{0}^{\pi/6}cos(x^2)~dx \geq sin~x \vert_{0}^{\pi/6}$
$\int_{0}^{\pi/6}cos(x^2)~dx \geq sin~\frac{\pi}{6} -sin~0$
$\int_{0}^{\pi/6}cos(x^2)~dx \geq \frac{1}{2}$
