Definate Integrals

**Truthbetold** · March 4th 2008, 08:37 PM

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?

The answer is 7/6.

**Mathnasium** · March 4th 2008, 09:27 PM

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!

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