We split [2, 4] into
subintervals of length
,
![[2,4]=\left[2,2+\dfrac2n\right]\cup\left[2+\dfrac2n,2+\dfrac4n\right]\cup\left[2+\dfrac4n,2+\dfrac6n\right]\cup\cdots\cup\left[2+\dfrac{2(n-1)}n,4\right]](https://tex.z-dn.net/?f=%5B2%2C4%5D%3D%5Cleft%5B2%2C2%2B%5Cdfrac2n%5Cright%5D%5Ccup%5Cleft%5B2%2B%5Cdfrac2n%2C2%2B%5Cdfrac4n%5Cright%5D%5Ccup%5Cleft%5B2%2B%5Cdfrac4n%2C2%2B%5Cdfrac6n%5Cright%5D%5Ccup%5Ccdots%5Ccup%5Cleft%5B2%2B%5Cdfrac%7B2%28n-1%29%7Dn%2C4%5Cright%5D)
so that the right endpoints are given by the sequence

for
. Then the Riemann sum approximating

is

The integral is given exactly as
, for which we get

To check: we have

Answer:
SI = $1,200 ; A = $6,200
Step-by-step explanation:
First, converting R percent to r a decimal
r = R/100 = 3%/100 = 0.03 per year,
then, solving our equation
I = 5000 × 0.03 × 8 = 1200
I = $ 1,200.00
The simple interest accumulated
on a principal of $ 5,000.00
at a rate of 3% per year
for 8 years is $ 1,200.00.
Answer:
y=-0.5x+8
Step-by-step explanation:
stat, or just rise over run
Answer:
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