$4-22$ Determine the shear and moment throughout the

tapered beam as a function of $x.$

$+{\displaystyle \underset{?}{\overset{?}{}}}{F}_y=0$;$,36-\frac{1}{2}(\frac{8}{9}x)(x)-\frac{8}{9}(8-\frac{8}{9}x)x-V=0$

$V=36-\frac{4}{9}{x}^2-8x+\frac{8}{9}{x}^2$

$V=0\mathrm{.}444x^2-8x+36$

$+{\displaystyle \underset{?}{\overset{?}{}}}M=0$;$,108+\frac{1}{2}(\frac{8}{9}x)(x)(\frac{2}{3}x)+\frac{8}{9}(8-\frac{8}{9}x)(x)(\frac{x}{2})-36x+M=0$

$M=-108-\frac{8}{27}{x}^3-4x^2+\frac{8}{18}{x}^3+36x$

$M=0\mathrm{.}148x^3-4x^2+36x-108$

Ans.

$*4-36.$ Determine the shear and moment in the tapered beam as a function of $x.$

$+T{F}_y=0$;$,36-\frac{1}{2}(\frac{8}{9}x)(x)-\frac{8}{9}(9-x)x-V=0$

$V=36-\frac{4}{9}{x}^2-8x+\frac{8}{9}{x}^2$

$V=0\mathrm{.}444x^2-8x+36$ Ans

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}M=0$

$108+\frac{1}{2}(\frac{8}{9}x)(x)(\frac{2}{3}x)+\frac{8}{9}(9-x)(x)(\frac{x}{2})-36x+M=0$

$M=-108-\frac{8}{27}{x}^3-4x^2+\frac{8}{18}{x}^3+36x$

$M=0\mathrm{.}148x^3-4x^2+36x-108$

Ans

Explain the yellow part. Where did the $\frac{8}{9}$ in the $\frac{8}{9}(8-8/9$x$)$ come from? I thought the area would be $(8-8/9$x$)*$x$.$ Explain the algebra behind it$.$ Explain the second one as well.