The signed area between the graph of $f\left(x\right)=3x^2-7x+2$ and the x-axis on$[0,4].$

The given function is$f\left(x\right)=3x^2-7x+2,$and the given interval$[0,4].$

On the basis of fundamental of calculus the area between the curve of the function

$f\left(x\right)=3x^2-7x+2$and x-axis on interval $[0,4]$is given by

$A={\int }_0^4(3x^2-7x+2)dx$

This area is the signed area between the curve of the function and x-axis on interval $[0,4].$

We calculate the value of the integral that is

$$\begin{gathered} A={\int }_0^{4}3x^{2}dx+{\int }_0^4-7xdx+{\int }_0^{4}2dx \\ \Rightarrow A={\left[\frac{3x^3}{3}\right]}_0^4-7{\left[\frac{x^2}{2}\right]}_0^4+2{\left[x\right]}_0^4\Rightarrow A=(4^3-0^3)-\frac{7}{2}(4^2-0^2)+2(4-0) \\ \Rightarrow A=64-56+8 \\ \Rightarrow A=16 \end{gathered}$$

Thus, the signed area between the curve of the function and x-axis on interval is shown as shaded portion in Figure-1 equals $16$units.

We not a portion of the area lies below the x-axis that are in above calculation is negative.

So exact area of between curve of the function and x-axis is when this area is considered a positive.

Now we find the exact value of area between the curve of the function and x-axis on interval $[0,4]$using graphing calculator.

Exact area: $A={\int }_0^4\left|3x^2-7x+2\right|dx$

The exact value of area between the curve of the function and x-axis on interval $[0,4]$is shown as shaded portion in Figure-2 equals $20.6296$units.

The signed area the signed area between the curves of the function is $16$units.