1. ## Definate Integrals

The inequality $sec x \geq 1 + \frac{x^2}{2}$ holds on (pi/x, -pi/2). Use it to find a lower bound for the value of $\int \sec x dx$on the interval [0,1].

(I need LaTex help with putting the numbers up on the integral)

I have no idea how to do this.
What do they mean by lower bound? How do we find that with a inequality like this?

2. The fact that $sec x \geq 1 + \frac{x^2}{2}$ on (0,1) means, basically, that if we find $\int_0^1{1 + \frac{x^2}{2}}dx$, our answer will be less than $\int secx$ on that same interval. Thinking about the integral as the area under the curve, if $sec x \geq 1 + \frac{x^2}{2}$, then the area under sec(x) on (0,1) must be greater than or equal to the area under $1 + \frac{x^2}{2}$.
$\int_0^1{1 + \frac{x^2}{2}}dx$, which gives
$x + \frac{x^3}{6}$ evaluated from 0 to 1, which gives your solution of $\frac{7}{6}$.