# Definate Integrals

• Mar 4th 2008, 08:37 PM
Truthbetold
Definate Integrals
The inequality $\displaystyle sec x \geq 1 + \frac{x^2}{2}$ holds on (pi/x, -pi/2). Use it to find a lower bound for the value of$\displaystyle \int \sec x dx$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?

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