Splitting up the interval of integration into 
subintervals gives the partition
![\left[0,\dfrac1n\right],\left[\dfrac1n,\dfrac2n\right],\ldots,\left[\dfrac{n-1}n,1\right]](https://tex.z-dn.net/?f=%5Cleft%5B0%2C%5Cdfrac1n%5Cright%5D%2C%5Cleft%5B%5Cdfrac1n%2C%5Cdfrac2n%5Cright%5D%2C%5Cldots%2C%5Cleft%5B%5Cdfrac%7Bn-1%7Dn%2C1%5Cright%5D)
Each subinterval has length 
. The right endpoints of each subinterval follow the sequence
with 
. Then the left-endpoint Riemann sum that approximates the definite integral is
and taking the limit as 
gives the area exactly. We have
Answer:
It would be 55.88 but if you round it it would be 55.9 then round it again it would be 56%.
Answer:
Help with what?
Step-by-step explanation:
I would be happy to answer but you need to attach a question.
The correct answer is 2 over 12 because you can simplify it
The apothem of this is 13
