Answer
$\int_0^1 \sec x dx \geq \dfrac{7}{6}$
Work Step by Step
When the function $f(x)$ has minimum and maximum on $[a,b]$ then, we have
$min f(b-a) \leq \int_a^b f(x) dx \leq max f(b-a)$
Given: $f(x)=\sec x \geq 1+\dfrac{x^2}{2}$ for $x \geq 0$
Now, we have
$ \int_0^1 \sec x dx \geq \int_0^1 (1+\dfrac{x^2}{2}) dx$
or, $ \int_0^1 \sec x dx \geq \int_0^1 dx +(1/2) \int_0^1 x^2 dx$
or, $\int_0^1 \sec x dx \geq 1+\dfrac{1}{6}$
Thus, $\int_0^1 \sec x dx \geq \dfrac{7}{6}$
