Definite integral |4-t| dt from x=3 to 7
using midpoint Riemann sums with the following partitions of the interval : [3,7]
(a) Partititioning into two nonequal subintervals [3,4] and [4,7] . Then the midpoint Riemann sum = ?
(b) Using 4 subintervals of equal length. Then the midpoint Riemann sum = ?
If anybody understands what I am talking about, help would be greatly appreciated. I understand it to be asking for the area and I came up with -4, but the site that I am doing the homework on tells me that I am wrong. Is it asking for something else or am I just wrong to think that it is -4.
Thanks
using midpoint Riemann sums with the following partitions of the interval : [3,7]
(a) Partititioning into two nonequal subintervals [3,4] and [4,7] . Then the midpoint Riemann sum = ?
(b) Using 4 subintervals of equal length. Then the midpoint Riemann sum = ?
If anybody understands what I am talking about, help would be greatly appreciated. I understand it to be asking for the area and I came up with -4, but the site that I am doing the homework on tells me that I am wrong. Is it asking for something else or am I just wrong to think that it is -4.
Thanks
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Since the integrand >=0 the area can never be negative.
(a) h1 = 4-3 = 1, x1 = (3+4)/2 = 3.5, h2 = 7-4 = 3, x2 = (4+7)/2 = 5.5.
sum = |4-x1| h1 + |4-x2| h2 = |4-3.5| * 1 + |4-5.5| * 3 = 5.
(b) h1=h2=h3=h4=1, x1=3.5, x2=4.5, x3 = 5.5, x4 = 6.5.
sum = |4-x1| h1 + |4-x2| h2 + |4-x3| h3 + |4-x4| h4
= 0.5*1 + 0.5*1 + 1.5*1 + 2.5*1
= 5.
The Riemann sum gives exact answer for this problem as long as one of the interval boundary is at x=4.
(a) h1 = 4-3 = 1, x1 = (3+4)/2 = 3.5, h2 = 7-4 = 3, x2 = (4+7)/2 = 5.5.
sum = |4-x1| h1 + |4-x2| h2 = |4-3.5| * 1 + |4-5.5| * 3 = 5.
(b) h1=h2=h3=h4=1, x1=3.5, x2=4.5, x3 = 5.5, x4 = 6.5.
sum = |4-x1| h1 + |4-x2| h2 + |4-x3| h3 + |4-x4| h4
= 0.5*1 + 0.5*1 + 1.5*1 + 2.5*1
= 5.
The Riemann sum gives exact answer for this problem as long as one of the interval boundary is at x=4.