The summand (R?) is missing, but we can always come up with another one.
Divide the interval [0, 1] into
subintervals of equal length
:
![[0,1]=\left[0,\dfrac1n\right]\cup\left[\dfrac1n,\dfrac2n\right]\cup\cdots\cup\left[1-\dfrac1n,1\right]](https://tex.z-dn.net/?f=%5B0%2C1%5D%3D%5Cleft%5B0%2C%5Cdfrac1n%5Cright%5D%5Ccup%5Cleft%5B%5Cdfrac1n%2C%5Cdfrac2n%5Cright%5D%5Ccup%5Ccdots%5Ccup%5Cleft%5B1-%5Cdfrac1n%2C1%5Cright%5D)
Let's consider a left-endpoint sum, so that we take values of
where
is given by the sequence

with
. Then the definite integral is equal to the Riemann sum




96 with a remainder of 5.
Answer:
254
Step-by-step explanation:
Answer:
Four billion, two-hundred seventy five thousand.
Answer: D 30%
Step-by-step explanation:amount change/amount start x 100%
(20-14)/20=.3
.3 x 100% = 30%