Understanding the Midpoint Rule for Integrals with n=6 and Homework Equations

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SUMMARY

The discussion focuses on applying the Midpoint Rule for calculating the integral of the function √(x²+1) from 2 to 14 using n=6 subintervals. The midpoints identified are 3, 5, 7, 9, 11, and 13, with each subinterval having a width of 2. The area under the curve is approximated by summing the areas of rectangles formed by these midpoints and the constant width. The formula used is 2 * f(x_i), where f(x_i) is the function evaluated at each midpoint.

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  • Understanding of integral calculus concepts, specifically the Midpoint Rule.
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  • Basic knowledge of how to calculate areas of rectangles.
  • Ability to perform arithmetic operations with real numbers.
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Homework Statement



Integral Upper Limit 14, Lower Limit 2, [squareroot(x^2+1)]dx ; n=6

Homework Equations





The Attempt at a Solution


Midpoints : 3,5,7,9,11,13
width of subintervals: (14-2)/6=2

2([squareroot(3^2+1)]+[squareroot(5^2+1)] ...


is that on the right track?
 
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Yes, this is the right track. You make it look so hard. You are adding up areas of rectangles right?

-Each rectangle has a height and a width.
-The height is determined by using some point on the interval that the rectangle is on, in this case, the midpoint. So you take the midpoint and plug it into the function to get the height.

Once you get that, all you need is the width which was 2. So just 2 * f(x_i) where each x_i is a midpoint.
 

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