Your bathtub has 12 gallons of water in it and you start to drain it. The water drains at a constant rate of 1/2
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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
Let's consider a simple example of estimating the value of a general definite integral,
Split up the interval
where
Now denote the left- and right-endpoint approximations by
So, you have
Now let
Factoring out
which is equivalent to
and is the average of
So the trapezoidal approximation for your problem should be