$$ x(t):=\begin{cases} 1&\text{if $0<t<T$}\\ \\ 0&\text{if otherwise} \end{cases} \qquad\text{and}\qquad h(t):=\begin{cases} t&\text{if $0<t<2T$}\\ \\ 0&\text{if otherwise} \end{cases} $$ Compute $(x*h)(t)$.
My approach: We first note that :$$\displaystyle x(\tau)=\begin{cases} 1&\text{if $0<\tau<T$}\\ \\ 0&\text{if otherwise} \end{cases}\qquad\text{and}\qquad\displaystyle h(t-\tau)=\begin{cases} t-\tau&\text{if $0<t-\tau<2T$}\\ \\ 0&\text{if otherwise} \end{cases}$$ We observe that : $$ 0<\tau<T $$ $$ t-2T<\tau<t $$ Thus, $(x*h)(t)\in\mathcal{R}[\max(0,t-2T),\min(t,T)]$ (i.e. Riemann integrable). Therefore : \begin{align*} (x*h)(t)&:=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)\;\text{d}\tau\\ &=\int_{\max(0,t-2T)}^{\min(t,T)}t-\tau\;\text{d}\tau\\ &=\left. t\tau-\frac{\tau^{2}}{2}\right|_{\max(0,t-2T)}^{\min(t,T)}\\ &=\left[t\min(t,T)-\frac{\min^{2}(t,T)}{2}\right]-\left[t\max(0,t-2T)-\frac{\max^{2}(0,t-2T)}{2}\right]\\ &=t[\min(t,T)-\max(0,t-2T)]-\frac{1}{2}[\text{min}^{2}(t,T)-\text{max}^{2}(0,t-2T)] \end{align*} How do I go from here?
You will probably have to discusses different cases according to the value of $t$.
It is slightly easier with the other paramatrization (or change of variable) and the same arguments (if there is no mistake) $$ (x\ast h)(t) = \int_{\mathbb{R}} x(t-\tau) \ h(\tau)\ d\tau = \int_{\max(t-T,0)}^{\min(t,2T)} x(t-\tau) \ h(\tau)\ d\tau = \int_{\max(t-T,0)}^{\min(t,2T)} \tau\ d\tau = \left[ \frac{\tau^2}{2}\right]_{\max(t-T,0)}^{\min(t,2T)} $$ One may use the Heaviside "function" $\theta$ (but this is in fact just a rewriting...) $$(x\ast h)(t) = \frac{\big(2\ T - (2T -t)\ \theta(2T -t) \big)^2}{2} - \frac{\big( (t-T)\ \theta (t-T) \big)^2}{2}$$ The result is a function defined "part" by "part" by simple polynomials so there has to be a discussion as for $x$ and $h$, if $t$ sthg, then sthg.
What you probably want is the usual interpretation of convolution. If we considered $\int_{\mathbb{R}^2} f(x)\ g(y)\ dx\ dy$, the domain of integration is the plan. Now if we have a parametrized curve $\gamma: \mathbb{R} \to \mathbb{R}^2$ we considered $\int_{\mathbb{R}} f\big(\gamma_1(t)\big)\ g\big(\gamma_2(t)\big)\ \lambda_{\gamma(t)}\big(\gamma'(t)\big) \ dt $. ($\gamma'\in \mathbb{R}^2$ is a vector so I introduced a linear form $\lambda$ that depends on the point where are at). The "domain" of integration now is the trajectory of the path and the convolution product can be understood as a particular case. In comparison to the first integral, $x$ is restricted to the first component $\gamma_1$ of $\gamma$ while $y$ is restricted to the second component. For a convolution product $(f\ast g)(t)$ and a given $t$, the trajectory is the subset $$\Gamma_t := \left\lbrace (x,y)\in \mathbb{R}^2,\ x\in \mathrm{supp}(f),\ y\in \mathrm{supp}(g),\ x+y =t\right\rbrace $$ but the condition $y=t-x$ (at $t$ fixed) in the plan $\mathbb{R}^2$ is precisely a "downward" line.
Let us look at the behavior of the integrand $f(x)\ g(y)$ with $(x,y)\in \Gamma_t$. If we moved only "horizontally" (only $x$ varies, doesn't happen if we want to stay in $\Gamma_t$), only $f$ varies and similarly if we move "vertically". A set of the form $x$ fixed, $y$ arbitrary corresponds to a vertical line and $\left\lbrace x\in \mathrm{supp}(f),\ y\in \mathbb{R} \right\rbrace $ to a vertical band. Support of $f(x)\ g(y)$ thus lies on a rectangle $\mathrm{supp}(f)\times \mathrm{supp}(g)$. For the convolution product, we thus have to look at the intersection $\Gamma_t$ with this rectangle.
Back to the example, $f$ is the function $x$ in the original post, just a cut-off function and $g$ is $h$ the unit monomial of order 1 (on a certain interval). With my parametrization, we put the support of $f:=x$ on the horizontal axis and that of $g:=h$ on the vertical axis. With my parametrization of the integral, the variable $x$ is "restricted" to $(t-\tau)$ while $y$ is "restricted" to $\tau$
So we should interpret $\tau$ as the vertical position, and just look at the portion given by the intersection of $\Gamma_t$ (downward line, 45°) with the rectangle (cartesian product of the supports), and integrate $\tau$ only for $\tau$ in the interval corresponding to vertical position corresponding to this portion. Sorry for this tedious description: make a drawing!!!