show that the left is $(8/60)*$wl$^3$ Continuous Beam UVL

$R_{\text{l}\text{a}\text{t}}$

$\frac{z}{l-x}=\frac{}{l}=>z=\frac{}{l}(l-x)$

$M_x=\frac{l}{3}x-\frac{}{l}l(-x)x**\frac{x}{2}+\frac{1}{2}[-\frac{(l-x)}{l}]x**\frac{2x}{3}$

$M_x=\frac{l}{3}x-\frac{l{x}^2}{2l}+\frac{x^3}{2l}-\frac{1}{2l}[l-l+x]x**\frac{2}{3}$

$M_x=\frac{lx}{3}-\frac{x^2}{2}+\frac{x^3}{2l}-\frac{x^3}{3l}$

$M_x=\frac{lx}{3}-\frac{x^2}{2}+\frac{x^3}{6l}$

function $f(x)=M_x$

$\frac{z}{l-x}=\frac{}{l}=>z=\frac{}{l}(l-x)$

$M_x=\frac{lx}{3}-\frac{(l-x)x**\frac{x}{l}}{l}+\frac{1}{2}\frac{-((l-x)]x}{l}$

Area $={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}f(x)dx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{}_{x}dx$

$A\stackrel{}{x}={\displaystyle \underset{0}{\overset{L}{}}}xf(x)dx$

$={\displaystyle \underset{0}{\overset{L}{}}}x{M}_{x}dx$

$A\stackrel{}{x}={\displaystyle \underset{0}{\overset{L}{}}}x[\frac{lx}{3}-\frac{x^2}{2}+\frac{x^3}{6l}]dx$

$={\displaystyle \underset{0}{\overset{L}{}}}(\frac{l{x}^2}{3}-\frac{x^3}{2}+\frac{x^4}{6l})dx$

$=\frac{l{x}^3}{9}-\frac{x^4}{8}+\frac{x^5}{30l}{|}_0^L||$

$A\stackrel{}{x}=\frac{L^4}{9}-\frac{L^4}{8}+\frac{l^4}{30}$

$=\frac{7}{360}{l}^4$

Now

$\frac{6A(\stackrel{}{x})}{l}=6**\frac{7}{360}\frac{l^4}{l}$

$=\frac{7}{60}{l}^3$

$\frac{\text{L}\text{E}\text{F}\text{T}}{\text{S}\text{h}\text{o}\text{w}\text{t}\text{h}\text{a}\text{t}\text{t}\text{i}\text{s}}=\frac{8}{60}{l}^3$

$\frac{?}{{?}_{E}FT}=,=\frac{8}{60}{l}^3$