Decimal form: -2.6666..
Fraction form: -2 2/3
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,
![\displaystyle\int_a^bf(x)\,\mathrm dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_a%5Ebf%28x%29%5C%2C%5Cmathrm%20dx)
Split up the interval
![[a,b]](https://tex.z-dn.net/?f=%5Ba%2Cb%5D)
into
![n](https://tex.z-dn.net/?f=n)
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
![a=x_0](https://tex.z-dn.net/?f=a%3Dx_0)
and
![b=x_n](https://tex.z-dn.net/?f=b%3Dx_n)
. Each subinterval has measure (width)
![\dfrac{a-b}n](https://tex.z-dn.net/?f=%5Cdfrac%7Ba-b%7Dn)
.
Now denote the left- and right-endpoint approximations by
![L](https://tex.z-dn.net/?f=L)
and
![R](https://tex.z-dn.net/?f=R)
, respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are
![\{x_0,x_1,\cdots,x_{n-1}\}](https://tex.z-dn.net/?f=%5C%7Bx_0%2Cx_1%2C%5Ccdots%2Cx_%7Bn-1%7D%5C%7D)
. Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints,
![\{x_1,x_2,\cdots,x_n\}](https://tex.z-dn.net/?f=%5C%7Bx_1%2Cx_2%2C%5Ccdots%2Cx_n%5C%7D)
.
So, you have
![L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)](https://tex.z-dn.net/?f=L%3D%5Cdfrac%7Bb-a%7Dn%5Cleft%28f%28x_0%29%2Bf%28x_1%29%2B%5Ccdots%2Bf%28x_%7Bn-2%7D%29%2Bf%28x_%7Bn-1%7D%29%5Cright%29)
![R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)](https://tex.z-dn.net/?f=R%3D%5Cdfrac%7Bb-a%7Dn%5Cleft%28f%28x_1%29%2Bf%28x_2%29%2B%5Ccdots%2Bf%28x_%7Bn-1%7D%29%2Bf%28x_n%29%5Cright%29)
Now let
![T](https://tex.z-dn.net/?f=T)
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,
![T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)](https://tex.z-dn.net/?f=T%3D%5Cdfrac%7Bb-a%7Dn%5Cleft%28%5Cdfrac%7Bf%28x_0%29%2Bf%28x_1%29%7D2%2B%5Cdfrac%7Bf%28x_1%29%2Bf%28x_2%29%7D2%2B%5Ccdots%2B%5Cdfrac%7Bf%28x_%7Bn-2%7D%29%2Bf%28x_%7Bn-1%7D%29%7D2%2B%5Cdfrac%7Bf%28x_%7Bn-1%7D%29%2Bf%28x_n%29%7D2%5Cright%29)
Factoring out
![\dfrac12](https://tex.z-dn.net/?f=%5Cdfrac12)
and regrouping the terms, you have
![T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)](https://tex.z-dn.net/?f=T%3D%5Cdfrac%7Bb-a%7D%7B2n%7D%5Cleft%28%28f%28x_0%29%2Bf%28x_1%29%2B%5Ccdots%2Bf%28x_%7Bn-2%7D%29%2Bf%28x_%7Bn-1%7D%29%29%2B%28f%28x_1%29%2Bf%28x_2%29%2B%5Ccdots%2Bf%28x_%7Bn-1%7D%29%2Bf%28x_n%29%29%5Cright%29)
which is equivalent to
![T=\dfrac12\left(L+R)](https://tex.z-dn.net/?f=T%3D%5Cdfrac12%5Cleft%28L%2BR%29)
and is the average of
![L](https://tex.z-dn.net/?f=L)
and
![R](https://tex.z-dn.net/?f=R)
.
So the trapezoidal approximation for your problem should be
It is 3 and 1/14ft or 3 and 4/16 ft taller
Answer:
1/5 hours i.e 0.2 hours (which is 12 mins)
Step-by-step explanation:
First note that 12 1/2 = 12.5, and 2 1/2 = 2.5.
To work out how many hours it takes hime to drive one mile, we divide the number of hours (2.5) by the number of miles (12.5),
i.e 2.5 ÷ 12.5
which equals 1/5 = 0.2 hours (= 12 mins).