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




Answer:
A = 8, B = 9 and C = 6.
Step-by-step explanation:
We can create equations to solve this:
6 + A = 14
So A = 8.
A + B = 17
8 + B = 17
B = 9.
C = 29 - (6 + A + B)
= 29 - ( 6 + 8 + 9)
= 29 - 23
= 6.