I have two equations $$x(t) = u(t) - 2u(t-2) + u(t-5)$$ $$h(t) = e^{2t}u(1-t)$$ where $u(t)$ is the unit step function.
I'm attempting to find the convolution of the two: $$y(t) = h(t)*x(t)$$
I decided that I would do the convolution of just u(t) and sum the transformations of that to get the total convolution. For this I get $$y(t) = \int^{\infty}_{-\infty}{u(\tau)e^{2(t-\tau)}u(1-t+\tau)d\tau}$$
I can break this into a piecewise function and remove u(t): $$y(t) = \cases{0 &\text{ if } t\lt 0\cr \int^{\infty}_{0}{e^{2(t-\tau)}u(1-t+\tau)d\tau} &\text{ if } t\ge 0}$$
It's at this point that I feel more on shaky ground. Looking at the remaining step function, I would say that $u(\tau)$ dominates when $t \ge 1$ and $u(t-1+\tau)$ dominates for $t \lt 1$
Thus I get a new expression... $$y(t) = \cases{\int^{\infty}_{0}{e^{2(t-\tau)d\tau}} &\text{ if } t\ge 1 \cr \int^{\infty}_{t-1}{e^{2(t-\tau)d\tau}} &\text{ if }0 \lt t\lt 1 \cr 0 &\text{ if } t\le 0}$$
From there I can do indefinite integration on each integral for a solution.
But I'm not sure that I developed the second piecewise function correctly...
$u(1-t+\tau) > 0$ for all $1-t + \tau \gt 0 \implies \tau \gt t-1$, but we also need $\tau \gt 0$.
Thus for $t \ge 1$ we have $\int_{t-1}^{\infty} \mathbb{e}^{2(t-\tau)} \,\mathbb d \tau$ and for $0 \lt t \lt 1$ we have $\int_{0}^{\infty} \mathbb{e}^{2(t-\tau)}\,\mathbb d \tau$