91Ó°ÊÓ

Diagram:

The area under the curve between each of the respective function and the $x-axis$is given by the definite integral on the interval $[a,b]as{∫}_a^{b}f\left(x\right)dx.$

The area under investigation of the sign W can be calculated as given below.

$A\left(x\right)={∫}_{x-5.29}^{x-18}r\left(x\right)dx-{∫}_{x-1.5}^{x-18}f\left(x\right)dx+{∫}_{x-18}^{x-30.71}s\left(x\right)dx{∫}_{x-18}^{x-34.49}g\left(x\right)dx$

where,

$$\begin{gathered} f(1.5)=55andf(22.49)=55 \\ r(5.29)=55andr(18.71)=55 \\ g(13.51)=55andg(34.49)=55 \\ s(17.29)=55ands(30.71)=55 \end{gathered}$$

Here,

$$\begin{gathered} {∫}_{x-5.29}^{x-18}r\left(x\right)dx \\ ={∫}_{x-5.29}^{x-18}{(x-12)}^2+10dx \\ ={∫}_{x-5.29}^{x-18}{x}^2-24x+154dx \\ ={[\frac{x^3}{3}-12x^2+154x]}_{5.29}^{18}=[1944-3888+2772-49.3+335.8-814.7] \\ =299.8 \end{gathered}$$

Now,

$$\begin{gathered} {∫}_{x-1.5}^{x-18}f\left(x\right)dx \\ ={∫}_{x-1.5}^{x-18}0.5{(x-12)}^{2}dx={∫}_{x-1.5}^{x-18}0.5(x^2-24x+144)dx \\ =0.5{[\frac{x^3}{3}-12x^2+144x]}_{1.5}^{18}=0.5[1944-3888+2592-1.125+27-216] \\ =0.5(457.875) \\ =228.94 \end{gathered}$$

And,

$$\begin{gathered} {∫}_{x-18}^{x-30.71}s\left(x\right)dx \\ ={∫}_{x-18}^{x-30.71}{(x-24)}^2+10dx \\ ={∫}_{x-18}^{x-30.71}{x}^2-48x+586dx \\ ={[\frac{x^3}{3}-24x^2+586x]}_{18}^{30.71}=[9654.2-22634.4+17996-1944+7776-10548] \\ =299.8 \end{gathered}$$

And,

$$\begin{gathered} {∫}_{x-18}^{x-34.49}g\left(x\right)dx \\ ={∫}_{x-18}^{x-34.49}0.5{(x-24)}^{2}dx={∫}_{x-18}^{x-34.49}0.5(x^2-24x+288)dx \\ =0.5{[\frac{x^3}{3}-12x^2+288x]}_{18}^{34.49}=0.5[13.675.98-14274.7+9933.12-1944+3888-5184] \\ =0.5\left[6094\right] \\ =3047.2 \end{gathered}$$

Since,

$$\begin{gathered} A\left(x\right)={∫}_{x-5.29}^{x-18}r\left(x\right)dx-{∫}_{x-1.5}^{x-18}f\left(x\right)dx+{∫}_{x-18}^{x-30.71}s\left(x\right)dx{∫}_{x-18}^{x-34.49}g\left(x\right)dx \\ A\left(x\right)=299.8-228.94+299.8-3047.2 \\ A\left(x\right)=-2676.54 \end{gathered}$$