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

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In summary, the integral with an upper limit of 14 and a lower limit of 2, using the function [squareroot(x^2+1)]dx and with n=6, can be solved by dividing the interval into 6 subintervals with a width of 2 and finding the area of each rectangle using the midpoints as the height and a width of 2.
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jkeatin
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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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  • #2
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.
 

What is the Midpoint Rule for Integrals?

The Midpoint Rule for Integrals is a numerical method for approximating the area under a curve, also known as an integral. It divides the region under the curve into smaller rectangles and calculates the area of each rectangle using the midpoint of each interval as the height. The sum of these areas gives an estimate of the integral.

How is the Midpoint Rule for Integrals calculated?

To use the Midpoint Rule for Integrals, the interval of integration is divided into subintervals of equal width. Then, the midpoint of each subinterval is calculated, and the corresponding rectangles are drawn. The area of each rectangle is found by multiplying the width of the subinterval by the height, which is the function evaluated at the midpoint. Finally, the sum of these areas gives an approximation of the integral.

What are the advantages of using the Midpoint Rule for Integrals?

The Midpoint Rule for Integrals is a relatively simple method that can provide a good estimate of the integral. It is also more accurate than the Left and Right Endpoint Rules, which use the endpoints of each subinterval as the height of the rectangles. Additionally, the Midpoint Rule can be easily programmed into a computer, making it useful for numerical integration in applications.

What are the limitations of the Midpoint Rule for Integrals?

While the Midpoint Rule for Integrals can provide a good estimate of the integral, it is not always accurate. The accuracy depends on the number of subintervals used and the shape of the curve. As the number of subintervals increases, the accuracy also improves. However, for curves with high curvature, a large number of subintervals may be needed to get a good approximation.

How is the error in the Midpoint Rule for Integrals calculated?

The error in the Midpoint Rule for Integrals can be estimated using the error bound formula, which takes into account the width of the subintervals and the maximum value of the second derivative of the function on the interval. The smaller the width of the subintervals and the smaller the maximum value of the second derivative, the smaller the error will be.

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