91Ó°ÊÓ

Consider the given question,

$A\left(x\right)$ is the function of the area between the graph of f and the x-axis from $0$ to x.

The $A\left(2.5\right)$is shaded in the diagram.

The function $A\left(x\right)$is becoming more greater on the interval $x∈\left[0,4\right]$.

Since, $A\left(0\right)<A\left(1\right)<A\left(2\right)<A\left(3\right)<A\left(4\right)$.

Then, the function $A\left(x\right)$is becoming more less on the interval $x∈[4,∞)$.

Since, $A\left(4\right)>A\left(5\right)>A\left(6\right)>A\left(7\right)>A\left(8\right)$.

Thus, function $A\left(x\right)$is increasing on $x∈\left[0,4\right]$and decreasing on$[4,∞)$.

For all $x∈\left[0,4\right]$, the function of $f\left(x\right)$is positive.

Since $f\left(x\right)â‰\yen 0$on the interval $x∈\left[0,4\right]$.

For all $x∈[4,∞)$, the function of $f\left(x\right)$is negative.

Since $f\left(x\right)≤0$on the interval $x∈[4,∞)$.

Thus, function of $f\left(x\right)$is positive on$x∈\left[0,4\right]$and negative on$x∈[4,∞)$.

As the function $A\left(x\right)$ of the area is defined between the graph of f and the x-axis from $0$ to x. Then,

$A\left(x\right)={∫}_0^{x}f\left(x\right)dx$

Working backwards below can be shown,

role="math" localid="1649830893032" $$\begin{gathered} \frac{dA\left(x\right)}{dx}=f\left(x\right) \\ dA\left(x\right)=f\left(x\right)dx \\ A\left(x\right)={∫}_0^{x}f\left(x\right)dx \end{gathered}$$

Thus,$A\text{'}\left(x\right)=f\left(x\right)$.