Definate Integrals

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 $1 + \frac{x^2}{2}$ .

So to find the lower bound, we really need to evaluate:

$\int_0^1{1 + \frac{x^2}{2}}dx$ , which gives

$x + \frac{x^3}{6}$ evaluated from 0 to 1, which gives your solution of $\frac{7}{6}$ .

Hope this helps!