Area functions, Find a formula for A(x)

Question

Let $A(x)$ be the area between the function and the $x$-axis and between the $y$-axis and the vertical line at a given $x$. Consider the following function.

$$f(t) = \begin{cases} - 2t + 8, & t \leqslant 3 \\ 2,& t > 3 \end{cases}$$

the textbook has the answer

$$A(x) = \begin{cases}8x - x^2, & 0 \leqslant x \leqslant 3 \\ 2x + 9,& {x > 3} \end{cases}$$

My rough work: area of triangle is $\frac 12 \cdot (-2x + 8)$, area of rectangle is $(2)(x-3)$.

Please explain how to get the answer.

Answer 1

Ok if you haven't heard of integration, you can solve this geometrically.

For the interval $[0,3]$ consider it as a trapeziod. Call this area $A_1$ the trapeziod for $x\leqslant3$ which given the defined function $f(t)$ will have area: $$A_1=\frac{1}{2}(b^1+b^2)\cdot l\tag{1}$$ Where $b^1$ and $b^2$ are the side lengths, defined by $f(t)$ and $l$ is the width, defined by $t$. So that expression $(1)$ becomes: $$A_1=\frac{1}{2}[(8) + (-2t+8)]t$$ $$=\frac{1}{2}(-2t+16)t$$ $$=\frac{1}{2}(-2t^2+16t)$$ $$=-t^2+8t$$ If your area function is defined in terms of x you now have: $$A(x)=8x-x^2, 0\leqslant x \leqslant3$$ For $x>3$ it appears the textbook takes into account the trapeziod shape. So the second part of the piecewise function is defined here as $2(x-3)+15$. The $15$ comes from $A(x)$ evaluated at $x = 3$. This expands to:

$$A(x)=2x + 9, x > 3$$ So as required you have your piecewise function: $$A(x) = \begin{cases}8x - x^2, & 0 \leqslant x \leqslant 3 \\ 2x + 9,& {x > 3} \end{cases}$$

Answer 2

Assuming the area under discussion is the area between the function $f(t)$, the $t$-axis, and the vertical lines $t=0$ and $t=x$.

$$\begin{align} f(t) & =\begin{cases}−2t+8 & : t\leqslant 3 \\[1ex] 2 & : t \ge 3\end{cases} \\[2ex]\text{by integration}\\[2ex] A(x) & =\begin{cases}\displaystyle \int_0^x 8-2 t \operatorname{d}t &: x\leqslant 3 \\[1ex]\displaystyle \int_0^3 -2t + 8 \operatorname{d}t +\int_3^x 2\operatorname{d}t & : x> 3 \end{cases} \\[1ex] & =\begin{cases} 8x-x^2 &: x\leqslant 3 \\[1ex] 8(3)-(3)^2 + 2x - 6 & : x> 3 \end{cases} \\[1ex] & =\begin{cases} 8x-x^2 &: x\leqslant 3 \\[1ex] 9 + 2x & : x> 3 \end{cases} \end{align}$$

If this is pre-calculus, then you simply break the figure into polygons as follows:

• For $x$ values of $x$ less than $3$ you have a rectangle of width $x$ and height $8-2x$, and a triangle of width $x$ and height $2x$

• For $x$ values greater than $3$ you have a rectangle of width $x$ and height $2$ plus a fixed triangle of base $3$ height $6$

Calculate this and the answer should be the same.