I am trying to find the lower and upper sum for a linear function, y=13-3x between x=0 and x=4.
To find this, I first have to find the upper and lower bound for each sub-interval.
I have found that the upper bound for the first subinterval is equal to 13. I found this by plugging in x=0 into the linear function, y=13-3x. This was found to be correct. However, when finding the upper bound for the second interval, all the way up to finding the upper bound for the fifth sub-interval, these answers were found to be wrong.
For example, to find the upper bound of the second interval, I simply plugged in x=1 into y=13-3x, resulting in 10, however, this was marked to be incorrect.
Any help will be appreciated. Thank you!
I would suppose that to actually answer the question would be to assume that you are considering a Riemann sum. If we consider the Riemann for the interval [0,4] then I would try to determine what Riemann sum we are evaluating here. I would imagine that since you stated that you have to find the lower and upper sum, we are going to have to find L{P;f} and U{P;f} where L and U are the lower and upper sum respectively. Where f is the function f(x)=y=13-3x
If we take P to be the partition of the interval into equal sub-intervals [0,1], [1,2], [2,3], and [3,4], then we already know that the lower sum will be what we call the right Riemann sum and the upper sum will be what we call the left Riemann sum. This is due to the fact that the function is decreasing.
Evaulating the left Riemann sum, we get
L{P;f} = 13+10+7+4 = 34
U{P;f} = 10+7+4+1 = 22
The answer will change based on what partition we are using. Different partitions will yield different answers, but in this case, I assumed the partition to be intervals of 1 unit each.
This is what I believe would be the answer to the question, but if it isn't please comment.
Since each sub-interval are to be equal and there are 5 sub-intervals, we now know that the partition is [0,4/5] [4/5,8/5] [8/5,12/5] [12/5, 16/5] [16/5,20/5].
Using the upper sum, we have,
f(0)*4/5 + f(4/5)*4/5 + f(8/5)*4/5 + f(12/5)*4/5 + f(16/5)*4/5 = 164/5
The lower sum,
f(4/5)*4/5 + f(8/5)*4/5 + f(12/5)*4/5 + f(16/5)*4/5 + f(20/5)*4/5 = 116/5