Split up the interval [0, 2] into <em>n</em> equally spaced subintervals:
![\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right]](https://tex.z-dn.net/?f=%5Cleft%5B0%2C%5Cdfrac2n%5Cright%5D%2C%5Cleft%5B%5Cdfrac2n%2C%5Cdfrac4n%5Cright%5D%2C%5Cleft%5B%5Cdfrac4n%2C%5Cdfrac6n%5Cright%5D%2C%5Cldots%2C%5Cleft%5B%5Cdfrac%7B2%28n-1%29%7Dn%2C2%5Cright%5D)
Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,
where 
. Each interval has length 
.
At these sampling points, the function takes on values of
We approximate the integral with the Riemann sum:
Recall that
so that the sum reduces to
Take the limit as <em>n</em> approaches infinity, and the Riemann sum converges to the value of the integral:
Just to check:
Answer:
Step-by-step explanation:
<u>The data given:</u>
- 93, 81, 94, 71, 89, 92, 94, 99
<u>Put the data in the ascending order:</u>
- 71, 81, 89, 92, 93, 94, 94, 99
<u>Since the data size is even, the median is the average of middle two:</u>
- median = (92 + 93)/2 = 92.5
Answer:
3pi/2 to 2π radians
Step-by-step explanation:
The centeal angle also has 295⁰. This value is between 275⁰ (or 3×pi/2) and 360⁰ (or 2×pi).
So, the answer is: 3pi/2 to 2π radians
