The trapezoidal approximation will be the average of the left- and right-endpoint approximations.
Let's consider a simple example of estimating the value of a general definite integral,

Split up the interval
![[a,b]](https://tex.z-dn.net/?f=%5Ba%2Cb%5D)
into

equal subintervals,
![[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]](https://tex.z-dn.net/?f=%5Bx_0%2Cx_1%5D%5Ccup%5Bx_1%2Cx_2%5D%5Ccup%5Ccdots%5Ccup%5Bx_%7Bn-2%7D%2Cx_%7Bn-1%7D%5D%5Ccup%5Bx_%7Bn-1%7D%2Cx_n%5D)
where

and

. Each subinterval has measure (width)

.
Now denote the left- and right-endpoint approximations by

and

, respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are

. Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints,

.
So, you have


Now let

denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

Factoring out

and regrouping the terms, you have

which is equivalent to

and is the average of

and

.
So the trapezoidal approximation for your problem should be
Answer:
64
Step-by-step explanation:
x + 5(x + 2)
Distribute
x + 5x+10
Combine like terms
6x+10
Let x = 9
6(9) +10
54+10
64
567 / 18 = 31.5
But because she can't make half a shelf, the answer would be 32 shelves
275,000 km
The moons A, B, C form a right triangle with AC forming a hypotenuse with the diameter of the planet and the distance of both moons from the surface. So add A to the surface, plus the diameter of the planet, plus surface to c.
115000+45000+115000 = 275000