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:
D i think sorry if wrong
Step-by-step explanation:
Answer:
D) x=5, x=-5.
Step-by-step explanation:
x^2-25=0
factor
(x+5)(x-5)=0
x+5=0, x-5=0,
x=0-5=-5,
x=0+5=5.
Answer:
or
Step-by-step explanation:
The given quadratic equation is

Group the constant terms on the right hand side.


Divide through by 5.

Add the square of half the coefficient of x., which is
to both sides of the equation.

The left hand side is now a perfect square.

Take the square root of both sides;



or 
or 
or 