Chapter 5.2, Problem 39E

To determine

The accumulator function $A\left(x\right)$ as a polynomial in x .

Answer to Problem 39E

The required accumulator function $A\left(x\right)$ as a polynomial of x is $A\left(x\right)=4x-x^2$ .

Explanation of Solution

Given data: The accumulator function in integral form is $A\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}\left(4-2t\right)dt,0\le }x\le 2$ .

Method/Formula used:

The integral $A\left(x\right)={\displaystyle \underset{a}{\overset{x}{\int }}f\left(t\right)dt,a\le }x\le b$ represents the area of the curve $y=f\left(t\right)$ , $a\le t\le b$ between $t=a$ and $t=x$ .

Calculation:

The accumulator function in integral form is $A\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}\left(4-2t\right)dt,0\le }x\le 2$ .

The area between the curve $f\left(t\right)=4-2t$ and t - axis from $t$ to $t=2$ is shown in Fig. 1 here.

  

From Fig. 1, the area $A\left(x\right)$ of the curve $y=f\left(t\right)$ and t - axis from $t=0$ to $t=x$ is equal to the area of the trapezium OPQRO given by

  $A\left(x\right)=\frac{1}{2}\left(OR+PQ\right)\cdot OP$ … (1)

Here, $OP=x$ , $PQ=f\left(x\right)$ that is equal to $4-2x$ and $OR=4$ so substitute $x$ for OP and $4-2x$ for PQ, and 4 for OR in result (1),

  $\begin{array}{c}A\left(x\right)=\frac{1}{2}\left(4+4-2x\right)\cdot x\\ =\frac{1}{2}\left(8-2x\right)\cdot x\\ =4x-x^2\end{array}$

Therefore, the accumulator function $A\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}\left(4-2t\right)dt,0\le }x\le 2$ as a polynomial of x is $A\left(x\right)=4x-x^2$ .