convolution of \$ e^{-t} \$ and 1-t
The functions to be convolved are \$h(t) \$ and \$u(t) \$, as shown in the top two diagrams in the figure below:

Figure source: I drew it.
The third diagram shows the folded \$u(t) \$, i.e., \$u(-t) \$ and the fourth diagram shows the folded and shifted \$u(t) \$, i.e., \$u(-t-\tau) \$. Note that \$\tau \$ is the shift variable. There will be two non-zero overlap integral scenarios: for Overlap A and Overlap B. These are schematically shown in the fifth and sixth diagrams in the figure.
The convolution is given by the following overlap integral:
$$ h(t) * u(t) = \int_{-\infty}^{+\infty}h(t) u(-t-\tau) dt $$
For the Overlap A scenario, the convolution integral reduces to:
$$ h(t) * u(t) = \int_{0}^{\tau}h(t) u(-t-\tau) dt $$
while for the Overlap B scenario, the convolution integral reduces to:
$$ h(t) * u(t) = \int_{\tau -1}^{\tau}h(t) u(-t-\tau) dt $$
Evaluation of convolution integral for Overlap A
$$\begin{align}
h(t)*u(t) &amp;= \int_0^\tau h(t) u(-t-\tau) dt \\
         &amp;= \int_0^\tau e^{-t}(1+t-\tau)dt    \\
         &amp;= (1-\tau) \int_0^\tau e^{-t}dt + \int_0^\tau te^{-t}dt \\
\end{align}$$
The two definite integrals are just special cases of well known indefinite integrals:
$$\int e^{ax}dx = e^{ax}/a$$ and
$$\int xe^{ax}dx = e^{ax}(ax - 1)/a^2$$
With \$a = -1\$ and \$x \$ replaced by \$t \$ in both integrals, we have
$$\int e^{-t}dt = -e^{-t}$$ and
$$\int te^{-t}dt = e^{-t}(-t - 1) = -e^{-t}(t + 1)$$
Thus, continuing the evaluation for Overlap A:
$$\begin{align}
h(t)*u(t) &amp;= (1-\tau) \left[-e^{-t} \right]_0^\tau + \left[e^{-t}(-t-1)    \right]_0^\tau \\
&amp;= (\tau -1)(e^{-\tau}-1)-[e^{-\tau}(\tau + 1) - 1]  \\
&amp;= \tau e^{-\tau}-e^{-\tau}-\tau+1-\tau e^{-\tau}-e^{-\tau}+1 \\
&amp;= 2-\tau -2e^{-\tau}
\end{align}$$
Replacing \$\tau \$ by \$t \$ then gives the desired result:
$$ y(t) = 2-t -2e^{-t} $$
Evaluation of convolution integral for Overlap B
$$\begin{align}
h(t)*u(t) &amp;= \int_{\tau -1}^\tau h(t) u(-t-\tau) dt \\
         &amp;= \int_{\tau -1}^\tau e^{-t}(1+t-\tau)dt    \\
         &amp;= (1-\tau) \int_{\tau -1}^\tau e^{-t}dt + \int_{\tau -1}^\tau te^{-t}dt \\
&amp;= (1-\tau) \left[-e^{-t} \right]_{\tau -1}^\tau + \left[e^{-t}(-t-1)    \right]_{\tau -1}^\tau \\
&amp;= (\tau -1)[e^{-\tau}-e^{-(\tau -1)}]-(\tau +1)e^{-\tau}+\tau e^{-(\tau -1)} \\
&amp;= \tau e^{-\tau}-e^{-\tau}-\tau e^{-(\tau -1)}+e^{-(\tau -1)}-\tau e^{-\tau}-e^{-\tau}+\tau e^{-(\tau -1)} \\
&amp;= e^{-(\tau -1)}-2e^{-\tau}
\end{align}$$
Replacing \$\tau \$ by \$t \$ then gives the desired result:
$$ y(t) = e^{-(t -1)}-2e^{-t} = e^{1 -t}-2e^{-t}  $$
Summary:
\$ y(t) = 0 \$ if \$t&lt;0\$
\$ y(t) = 2-t-2e^{-t} \$ if \$0\le t\le 1\$

\$ y(t) = e^{1-t}-2e^{-t} \$ if \$t&gt;1\$
Check:
Area under \$h(t) = \int_0^\infty h(t) dt = 1    \$. Area under \$u(t) = \int_0^1 u(t) dt = 1/2    \$. Area under \$y(t) \$ is
$$ \int_0^\infty y(t) dt = \int_0^1 (2-t -2e^{-t}) dt + \int_1^\infty (e^{1 -t}-2e^{-t}) dt = (\frac{2}{e} -\frac{1}{2})+(1-\frac{2}{e}) 
= \frac{1}{2} $$
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The alternative convolution option
Suppose it is desired to fold and shift \$h(t) \$ instead of \$u(t) \$. Then the first figure is replaced by this figure:

Figure source: I drew it.
The functions to be convolved are \$h(t) \$ and \$u(t) \$, as shown in the top two diagrams in the figure. The third diagram shows the folded \$h(t) \$, i.e., \$h(-t) \$ and the fourth diagram shows the folded and shifted \$h(t) \$, i.e., \$h(-t-\tau) \$. Note that \$\tau \$ is the shift variable. There will be two non-zero overlap integral scenarios: for Overlap A and Overlap B. These are schematically shown in the fifth and sixth diagrams in the figure.
The convolution is given by the following overlap integral:
$$ h(t) * u(t) = \int_{-\infty}^{+\infty}h(-t-\tau) u(t) dt $$
For the alternative Overlap A scenario, the convolution integral reduces to:
$$ h(t) * u(t) = \int_{0}^{\tau}h(-t-\tau) u(t) dt $$
while for the alternative Overlap B scenario, the convolution integral reduces to:
$$ h(t) * u(t) = \int_0^1 h(-t-\tau) u(t) dt $$
Evaluation of convolution integral for Overlap A
$$\begin{align}
h(t)*u(t) &amp;= \int_0^\tau h(-t-\tau) u(t) dt \\
         &amp;= \int_0^\tau e^{t-\tau} (1-t)dt    \\
&amp;= e^{-\tau} \left[\int_0^{\tau}e^t dt - \int_0^{\tau}te^t dt  \right]  \\
&amp;= e^{-\tau} \left[[e^t]_0^{\tau} - [e^t (t-1)]_0^{\tau}   \right]  \\
&amp;= e^{-\tau} \left[e^{\tau} -1 -\tau e^{\tau} + e^{\tau} -1   \right]  \\
&amp;= e^{-\tau} \left[2e^{\tau} -\tau e^{\tau} -2    \right] \\
         &amp;= 2-\tau -2e^{-\tau} \\
\end{align}$$
Replacing \$\tau \$ by \$t \$ then gives the desired result:
$$ y(t) = 2-t -2e^{-t} $$
Evaluation of convolution integral for Overlap B
$$\begin{align}
h(t)*u(t) &amp;= \int_0^1 h(-t-\tau) u(t) dt \\
         &amp;= \int_0^1 e^{t-\tau} (1-t)dt    \\
         &amp;= e^{-(\tau -1)}-2e^{-\tau} \\
\end{align}$$
Replacing \$\tau \$ by \$t \$ then gives the desired result:
$$ y(t) = e^{-(t -1)}-2e^{-t} = e^{1 -t}-2e^{-t}  $$