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Math Help - Definate Integrals

  1. #1
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    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?

    The answer is 7/6.
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  2. #2
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    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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