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:
let the solution be 
let the solutions be a = 3, b = 1
is the equation with consistent value.
<span>Let n = the number of nickles
Let q = the number of quarters
Then for your problem we have
(1) n + q = 43 and
(2) 5*n + 25*q = 100*6.95 (always work in cents to avoid decimal numbers) or
(3) 5*n + 25*q = 695
Now substitute n of (1) into (3) and get
(4) 5*(43 - q) + 25*q = 695 or
(5) 215 - 5*q + 25*q = 695 or
(6) 20*q = 695 - 215 or
(7) 20*q = 480 or
(8) q = 24
Then using (1) we get
(9) n + 24 = 43 or
(10) n = 19
Let's check these values.
Is (.05*19 + .25*24 = 6.95)?
Is (.95 + 6.00 = 6.95)?
Is (6.95 = 6.95)? Yes
Answer: Kevin and Randy have 19 nickles and 24 quarters in the jar.</span>
E) distributive property
Because you are distributing the 3 to 2/5x and 7/5y
