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:
-9
Step-by-step explanation:
The applicable exponent rule is ...
a^b·a^c = a^(b+c)
___
For the y factor, ...
y^-13 · y^4 = y^(-13+4) = y^-9 . . . . . exponent of -9
Answer:
Step-by-step explanation:
First consider the diagram:
Now, we know that
Now, the area of the triangle ABC is given by,
