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:
*3545*
Step-by-step explanation:
52 square + 29 square = 3545
then square root 3545 = 59.539 ( round it if you want)
Answer: 2343 / 256
Explanation
I will do this for you in two forms: 1) adding each term, and 2) using the general formula for the sum of geometric series.
1) Adding the terms:
4
∑ 3 (3/4)^i = 3 (3/4)^0 + 3 (3/4)^1 + 3 (3/4)^2 + 3 (3/4)^3 + 3 (3/4)^4
i=0
= 3 + 9/4 + 27/16 + 81/64 + 243/256 = [256*3 + 27*16 + 64*9 + 4*81 + 243] / 256 =
= 2343 / 256
2) Using the formula:
n-1
∑ A (r^i) = A [1 - r^(n) ] / [ 1 - r]
i=0
Here n - 1 = 4 => n = 5
r = 3/4
A = 3
Therefore the sum is 3 [ 1 - (3/4)^5 ] / [ 1 - (3/4) ] =
= 3 [ 1 - (3^5) / (4^5) ] / [ 1/4 ] = 3 { [ (4^5) - (3^5) ] / (4^5) } / {1/4} =
= (3 * 781) / (4^5) / (1/4) = 3 * 781 / (4^4) = 2343 / 256
So, no doubt, the answer is 2343 / 256
The answer is the fourth one. Trust me I had the same question
The area in which the right angle was at before. the angle nor the triangle changes