Generally we use n as a counter and begin it with 1: the first term is obtained when n = 1. Thus, n must be a positive integer, 1 or greater: {1, 2, 3, ... }.
Answer:
2.2
Step-by-step explanation:
Answer: her monthly payments would be $267
Step-by-step explanation:
We would apply the periodic interest rate formula which is expressed as
P = a/[{(1+r)^n]-1}/{r(1+r)^n}]
Where
P represents the monthly payments.
a represents the amount of the loan
r represents the annual rate.
n represents number of monthly payments. Therefore
a = $12000
r = 0.12/12 = 0.01
n = 12 × 5 = 60
Therefore,
P = 12000/[{(1+0.01)^60]-1}/{0.01(1+0.01)^60}]
12000/[{(1.01)^60]-1}/{0.01(1.01)^60}]
P = 12000/{1.817 -1}/[0.01(1.817)]
P = 12000/(0.817/0.01817)
P = 12000/44.96
P = $267
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:
