Estimating the value of an integral

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The discussion focuses on estimating the value of an integral using the properties of upper and lower bounds. Initially, there was confusion about the notation used for M and m, which led to incorrect attempts at the solution. The correct approach involves identifying the maximum and minimum values of the function 5/(1+x^2) over the interval [0,1]. The maximum occurs at x=0 and the minimum at x=1, allowing for the proper application of the inequality. Ultimately, the integral value is estimated correctly by applying these bounds.
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Homework Statement


It is known that if m[PLAIN]http://www.webassign.net/images/lteq.giff(x)[ATTACH=full]196966[/ATTACH]M for a[PLAIN]http://www.webassign.net/images/lteq.gifx[ATTACH=full]196967[/ATTACH]b, then the following property of integrals is true.

symimage.cgi?expr=%20m%28b-a%29%20%3C%3D%20int_a%5Eb%20f%28x%29dx%20%3C%3D%20M%28b-a%29%20.gif


Use this property to estimate the value of the given integral.

____________ <=
symimage.cgi?expr=%20int_0%5E1%205%2F%281%2Bx%5E2%29dx%20.gif
<= _____________

Homework Equations


The Attempt at a Solution


Honestly, I don't really know what it wants for a solution. I put in two numbers that satisfied the inequality, but it marked it wrong, so I assume it wants some sort of expression. The value of the integral itself is 5pi/4, but other than that I am at a complete loss for how to even begin.

Nevermind, I figured it out... I was overcomplicating it. I didn't realize they were using M and m as a function... I thought they were multiplying it. So m(1) <= 5/(1+1^2) = 5/2, and so-forth.
 

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They were multiplying it. Over the interval [0,1], the function 5/(1+x2) attains a maximum at x=0 and a minimum at x=1, so you can use the values m=f(1) and M=f(0) in the inequality.
 
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