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:
(C)−3 ≤ x ≤ 1
Step-by-step explanation:
The average rate of change of function h over the interval
, is given by this expression:

Given the function
on the interval:− 3 ≤ x ≤ 1

The average rate of change:

Therefore, the function has a negative average rate of change over the interval − 3 ≤ x ≤ 1.
CHECK:
(A)Average rate of change of h(x) over the interval − 3 ≤ x ≤ 5=2
(B)Average rate of change of h(x) over the interval 1 ≤ x ≤ 4=5
(D)Average rate of change of h(x) over the interval − 1 ≤ x ≤ 5=4
Baso 6 divided by 2 is 3 that’s what the question is saying
By PEMDAS (or any alternative), <em>the first step should be to divide</em>, as D comes before A. So, 16 ÷ 4.
Answer:y=-4
Step-by-step explanation: