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:
The first five terms of the sequence are 1, 4, 7, 10, 13.
Solution:
Given data:
General term of the arithmetic sequence.
, where d is the common difference.
d = 3
Put n = 2 in , we get
Put n = 3 in , we get
Put n = 4 in , we get
Put n = 5 in , we get
The first five terms of the sequence are 1, 4, 7, 10, 13.
Answer:
A. 50°
Step-by-step explanation:
The external angle ACB created by tangents CA and CB is the supplement of arc AB it intercepts.
∠ACB = 180° -AB
∠ACB = 180° -130°
∠ACB = 50°
if you're trying to solve for x, and it is equal to zero, then it would be -3. plug it in and it would all cancel out
Answer:
Step-by-step explanation:
Δ = 
VΔ = 4 * 8 = 32
a = (- 4 ± 32) * 
a' = 7/2
a''= 9/2
