Impulse response of a continuous time system Question
  
  The impulse response of a continuous time system is $h(t) = e^{-t}u(t)$. When $x(t) = u(t - 1)$ is input to this system what is
  the value of the output signal $y(t)$ at $t = 2.34$.


My Attempt

For convolution $y(t) = h(t) \star x(t)$
\begin{align}
x(2.27) &amp;= u(2.34-1) = 1\\
h(2.27) &amp;= e^{-2.34}\cdot u(2.34) = 0.09632763823\\
\Longrightarrow y(t) &amp;= 1 \cdot 0.09632 = 0.096
\end{align}

But the answer is wrong. Not too sure where my mistake is. Could someone correct my error?

Second Attempt
\begin{align}
y(t) &amp;= ∫x(\tau)h(t-\tau)d\tau \quad\text{with limits negative and positive infinity}\\
&amp;= ∫u(\tau-1)e^{-\tau}u(t-\tau)d\tau \\
&amp;= ∫e^{-\tau}u(\tau-1)u(t-\tau)d\tau \\
&amp;= ∫e^{-\tau}d\tau \quad \text{with limits 1 to 2.34}\\
&amp;= 0.2715518029 
\end{align}

Though the answer still appears to be wrong.
\begin{align}
y(2.34) &amp;= h \star x\,\big\vert_{t=2.34}\\
&amp;= \int_{-\infty}^\infty h(\tau)x(2.34-\tau)\,\mathrm d\tau
&amp;\scriptstyle{\text{from the definition of the convolution integral}} \\
&amp;= \int_0^\infty e^{-\tau}x(2.34-\tau)\,\mathrm d\tau &amp;\scriptstyle{\text{because } h(\tau)=0\text{ for }\tau &lt; 0}\\
&amp;= \int_0^\infty e^{-\tau}u(1.34-\tau)\,\mathrm d\tau &amp;\scriptstyle{\text{because }x(2.34-\tau) = u((2.34-\tau)-1)}\\
&amp;= \int_0^{1.34} e^{-\tau}\,\mathrm d\tau
&amp;\scriptstyle{\text{because }u(1.34-\tau) = 0 \text{ for }\tau &lt; 1.34}\\
&amp;= 1-e^{-1.34}
\end{align}