Splitting up [0, 3] into
equally-spaced subintervals of length
gives the partition
![\left[0, \dfrac3n\right] \cup \left[\dfrac3n, \dfrac6n\right] \cup \left[\dfrac6n, \dfrac9n\right] \cup \cdots \cup \left[\dfrac{3(n-1)}n, 3\right]](https://tex.z-dn.net/?f=%5Cleft%5B0%2C%20%5Cdfrac3n%5Cright%5D%20%5Ccup%20%5Cleft%5B%5Cdfrac3n%2C%20%5Cdfrac6n%5Cright%5D%20%5Ccup%20%5Cleft%5B%5Cdfrac6n%2C%20%5Cdfrac9n%5Cright%5D%20%5Ccup%20%5Ccdots%20%5Ccup%20%5Cleft%5B%5Cdfrac%7B3%28n-1%29%7Dn%2C%203%5Cright%5D)
where the right endpoint of the
-th subinterval is given by the sequence

for
.
Then the definite integral is given by the infinite Riemann sum

Answer:
the first and last ones
Step-by-step explanation:
Answer:
0.375
Step-by-step explanation:
= 3/8
= 3 ÷ 8
= 0.375
Answer:
Its a straight line
Step-by-step explanation: