The summand (R?) is missing, but we can always come up with another one.
Divide the interval [0, 1] into 
subintervals of equal length 
:
![[0,1]=\left[0,\dfrac1n\right]\cup\left[\dfrac1n,\dfrac2n\right]\cup\cdots\cup\left[1-\dfrac1n,1\right]](https://tex.z-dn.net/?f=%5B0%2C1%5D%3D%5Cleft%5B0%2C%5Cdfrac1n%5Cright%5D%5Ccup%5Cleft%5B%5Cdfrac1n%2C%5Cdfrac2n%5Cright%5D%5Ccup%5Ccdots%5Ccup%5Cleft%5B1-%5Cdfrac1n%2C1%5Cright%5D)
Let's consider a left-endpoint sum, so that we take values of 
where 
is given by the sequence
with 
. Then the definite integral is equal to the Riemann sum
Answer: x ( y + 1) + y ( y + 3) + 2
Step-by-step explanation:
( x + y + 2) ( y + 1)
x ( y + 1) + y ( y + 1) + 2 ( y + 1)
expanding the bracket
( xy + x) + ( y2 + y) + ( 2y + 2)
removing the bracket
xy + x + y2 +y + 2y + 2
collecting like terms
xy + x + y2 + 3y + 2
given as: x(y + 1) + y ( y + 3) + 2
Answer:– 1 – cos(ϴ)
Step-by-step explanation:
Answer:
Step-by-step explanation:
since there is no table included with the question I'm concluding that you need to find values of y using values of x, simply substitute the x values in the equation and you'll get the y value like this
if x = 1 y = 4
