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

[jkeatin]

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?

 

[mistermath]

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.