Using Laplace transform in $t$, or otherwise, solve the equation for $u$:
$$\frac{\partial u}{\partial t}+\frac{\partial u}{\partial x}=x$$
in the region: $x$ > 0, $t$ > 0, subject to the boundary condition $u$(0,t) = 0 and to the initial condition $u$(x,0)=$f$(x)
On
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \int_{0}^{\infty}\bracks{% \partiald{{\rm u}\pars{x,t}}{t} + \partiald{{\rm u}\pars{x,t}}{x}}\expo{-st} \,\dd t & = \int_{0}^{\infty}x\expo{-st}\,\dd t \\[3mm] \imp -\ \overbrace{{\rm u}\pars{x,0}}^{\ds{=\ {\rm f}\pars{x}}}\ + s\ \overbrace{\int_{0}^{\infty}{\rm u}\pars{x,t}\expo{-st}\,\dd t} ^{\ds{\equiv\ \hat{\rm u}\pars{x,s}}}\ + \partiald{}{x}\overbrace{\int_{0}^{\infty}{\rm u}\pars{x,t}\expo{-st}\,\dd t}^{\ds{\equiv\ \hat{\rm u}\pars{x,s}}}\ & =\ {x \over s} \end{align}
Then, $$ \pars{\partiald{}{x} + s}\hat{\rm u}\pars{x,s} = {x \over s} + {\rm f}\pars{x} \quad\imp\quad \partiald{\bracks{\expo{sx}\hat{\rm u}\pars{x,s}}}{x} = {x \over s}\,\expo{sx} + {\rm f}\pars{x}\expo{sx} $$
$$ \expo{sx}\hat{\rm u}\pars{x,s}\ -\ \overbrace{\hat{\rm u}\pars{0,s}}^{\ds{=\ 0}} = {1 \over s}\,{1 + \expo{sx}\pars{sx - 1} \over s^{2}} + \int_{0}^{x} {\rm f}\pars{x'}\expo{sx'}\,\dd x' $$
$$ \hat{\rm u}\pars{x,s} = \ {\expo{-sx} + sx - 1 \over s^{3}} + \expo{-sx}\int_{0}^{x} {\rm f}\pars{x'}\expo{sx'}\,\dd x' $$
\begin{align} {\rm u}\pars{x,t} & = \int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic} {\expo{s\pars{t - x}} \over s^{3}}\,{\dd s \over 2\pi\ic} + x\int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic} {\expo{st} \over s^{2}}\,{\dd s \over 2\pi\ic} - \int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic} {\expo{st} \over s^{3}}\,{\dd s \over 2\pi\ic} + \mbox{} \\[3mm] & \int_{0}^{x} {\rm f}\pars{x'}\bracks{% \int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic}\expo{s\pars{x' + t - x}}\,{\dd s \over 2\pi\ic}}\,\dd x' \\[5mm] & = \Theta\pars{t - x}\,\half\pars{t - x}^{2} + xt - \half\,t^{2} + \int_{0}^{x}{\rm f}\pars{x'}\delta\pars{x' + t - x}\,\dd x' \\[3mm] & = \Theta\pars{t - x}\,\half\pars{t - x}^{2} + xt - \half\,t^{2} + \Theta\pars{x - t}{\rm f}\pars{x - t} \end{align}
Method of characteristics
On any line $\mathrm{d}x=\mathrm{d}t$ (that is, $x=x_0+t$), we get $$ \mathrm{d}u=\frac{\partial u}{\partial t}\mathrm{d}t+\frac{\partial u}{\partial x}\mathrm{d}x=x\,\mathrm{d}x\tag{1} $$ Integrating, we get $$ u(x,x-x_0)=\frac12x^2+v(x_0)\tag{2} $$ or $$ u(x,t)=\frac12x^2+v(x-t)\tag{3} $$ for any differentiable function $v$.
Initial Conditions
The two conditions meet at $(0,0)$. The characteristic that passes through this point is $x=t$.
For $x\le t$, we use the initial condition $u(0,t)=0$, which means $$ \begin{align} v(0-t) &=u(0,t)-\frac12\cdot0^2\\ &=0\tag{4} \end{align} $$ For $t\gt x$, we use the initial condition $u(x,0)=f(x)$, which means $$ \begin{align} v(x-0) &=u(x,0)-\frac12x^2\\ &=f(x)-\frac12x^2\tag{5} \end{align} $$ Therefore, $$ v(x)=\left\{\begin{array}{} 0&\text{if }x\le0\\ f(x)-\frac12x^2&\text{if }x\gt0 \end{array}\right.\tag{6} $$ and we plug $(6)$ into $(3)$.