Answer:

Step-by-step explanation:
We want to find the Riemann sum for
with n = 6, using left endpoints.
The Left Riemann Sum uses the left endpoints of a sub-interval:

where
.
Step 1: Find 
We have that 
Therefore, 
Step 2: Divide the interval
into n = 6 sub-intervals of length 
![a=\left[0, \frac{\pi}{8}\right], \left[\frac{\pi}{8}, \frac{\pi}{4}\right], \left[\frac{\pi}{4}, \frac{3 \pi}{8}\right], \left[\frac{3 \pi}{8}, \frac{\pi}{2}\right], \left[\frac{\pi}{2}, \frac{5 \pi}{8}\right], \left[\frac{5 \pi}{8}, \frac{3 \pi}{4}\right]=b](https://tex.z-dn.net/?f=a%3D%5Cleft%5B0%2C%20%5Cfrac%7B%5Cpi%7D%7B8%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B%5Cpi%7D%7B8%7D%2C%20%5Cfrac%7B%5Cpi%7D%7B4%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B%5Cpi%7D%7B4%7D%2C%20%5Cfrac%7B3%20%5Cpi%7D%7B8%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B3%20%5Cpi%7D%7B8%7D%2C%20%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B%5Cpi%7D%7B2%7D%2C%20%5Cfrac%7B5%20%5Cpi%7D%7B8%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B5%20%5Cpi%7D%7B8%7D%2C%20%5Cfrac%7B3%20%5Cpi%7D%7B4%7D%5Cright%5D%3Db)
Step 3: Evaluate the function at the left endpoints






Step 4: Apply the Left Riemann Sum formula


Answer:
Because it increases the risk of Type 1 error
Step-by-step explanation:
ANOVA is the analysis of the variance .
When comparing more than two treatment means we use ANOVA because a t test increases the risk of type 1 error .
For example if we wish to compare 4 population means there will be 4C2 = 6 separate pairs and to test the null hypothesis that all four population means are equal would require six two sample t test. Similarly to test 10 population mean would require 45 separate two sample t test.
This has two disadvantages .
First the procedure is too lengthy and tediuos.
Second the overall level of significance greatly increases as the number of t- tests increases.
The analysis of the variance compares two different estimates of variance using the F distributionto determine whether the population means are equal.
Answer:
2
Step-by-step explanation:

Just think of slope as the rate of change