Is there a convolution mistake in my method? I have the input signal $x(t)$ 

And impulse response $h(t)=20 e^{-1000t} u(t)$ in which u(t) is the unit step function.

When I try a convolution, I thought the solutions would be something like:

$
\begin{array}{ll}
  \int\limits_{0}^{t} 2\cdot 20e^{-1000(t-\tau)}d\tau = \frac{1}{25}(1-e^{-1000t}),&amp;  0 \le t &lt;  2 \\
  \int\limits_{0}^{2} 2\cdot 20e^{-1000(t-\tau)}d\tau - \int\limits_{2}^{t}  20e^{-1000(t-\tau)}d\tau = \frac{3}{50}e^{2000-1000t}-\frac{e^{-1000t}}{25}-\frac{1}{50},&amp; 2\le t &lt;  3 \\
 \int\limits_{0}^{2} 2\cdot 20e^{-1000(t-\tau)}d\tau - \int\limits_{2}^{3}  20e^{-1000(t-\tau)}d\tau = \frac{3}{50}e^{2000-1000t}-\frac{1}{50}e^{3000-1000t}-\frac{e^{-1000t}}{25},&amp;  t \geq 3 \\
 \end{array}
$

But this did  not look like the answer in the solution manual, so did I do something wrong here?

Answer in solution manual: 

Alternative solution in solution manual:
You just multiplied the two functions and integrated them but you didn't convolve them. You must compute

$$y(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau=2\int_0^2h(t-\tau)d\tau-\int_2^3h(t-\tau)d\tau$$

Alternatively, you can compute the step response

$$a(t)=\int_{-\infty}^th(\tau)d\tau$$

Because $x(t)=2u(t)-3u(t-2)+u(t-3)$ the output can be written in terms of the step response:

$$y(t)=2a(t)-3a(t-2)+a(t-3)$$