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

$$begin{align} h(t)*u(t) &= int_0^tau h(t) u(-t-tau) dt \ &= int_0^tau e^{-t}(1+t-tau)dt \ &= (1-tau) int_0^tau e^{-t}dt + int_0^tau te^{-t}dt \ end{align}$$

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:

$$begin{align} h(t)*u(t) &= (1-tau) left[-e^{-t} right]_0^tau + left[e^{-t}(-t-1) right]_0^tau \ &= (tau -1)(e^{-tau}-1)-[e^{-tau}(tau + 1) - 1] \ &= tau e^{-tau}-e^{-tau}-tau+1-tau e^{-tau}-e^{-tau}+1 \ &= 2-tau -2e^{-tau} end{align}$$

Replacing $tau $ by $t $ then gives the desired result:

$$ y(t) = 2-t -2e^{-t} $$

Evaluation of convolution integral for Overlap B

$$begin{align} h(t)*u(t) &= int_{tau -1}^tau h(t) u(-t-tau) dt \ &= int_{tau -1}^tau e^{-t}(1+t-tau)dt \ &= (1-tau) int_{tau -1}^tau e^{-t}dt + int_{tau -1}^tau te^{-t}dt \ &= (1-tau) left[-e^{-t} right]_{tau -1}^tau + left[e^{-t}(-t-1) right]_{tau -1}^tau \ &= (tau -1)[e^{-tau}-e^{-(tau -1)}]-(tau +1)e^{-tau}+tau e^{-(tau -1)} \ &= tau e^{-tau}-e^{-tau}-tau e^{-(tau -1)}+e^{-(tau -1)}-tau e^{-tau}-e^{-tau}+tau e^{-(tau -1)} \ &= e^{-(tau -1)}-2e^{-tau} end{align}$$

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 $t<0$
$ y(t) = 2-t-2e^{-t} $ if $0le tle 1$
$ y(t) = e^{1-t}-2e^{-t} $ if $t>1$

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

$$begin{align} h(t)*u(t) &= int_0^tau h(-t-tau) u(t) dt \ &= int_0^tau e^{t-tau} (1-t)dt \ &= e^{-tau} left[int_0^{tau}e^t dt - int_0^{tau}te^t dt right] \ &= e^{-tau} left[[e^t]_0^{tau} - [e^t (t-1)]_0^{tau} right] \ &= e^{-tau} left[e^{tau} -1 -tau e^{tau} + e^{tau} -1 right] \ &= e^{-tau} left[2e^{tau} -tau e^{tau} -2 right] \ &= 2-tau -2e^{-tau} \ end{align}$$

Replacing $tau $ by $t $ then gives the desired result:

$$ y(t) = 2-t -2e^{-t} $$

Evaluation of convolution integral for Overlap B

$$begin{align} h(t)*u(t) &= int_0^1 h(-t-tau) u(t) dt \ &= int_0^1 e^{t-tau} (1-t)dt \ &= e^{-(tau -1)}-2e^{-tau} \ end{align}$$

Replacing $tau $ by $t $ then gives the desired result:

$$ y(t) = e^{-(t -1)}-2e^{-t} = e^{1 -t}-2e^{-t} $$

Writing down convolution product results in

$$ (h * u)(t) = int_{-infty}^{+infty}h(t - v)cdot u(v) dv $$

The parts of the domain we're interested in are:

$$t-vge 0 Rightarrow v leq t$$ for $h(t-v)$

$$0 leq v leq 1$$ for $u(v)$

We can then find that for $0 leq t leq 1$:

$$begin{align} (h*u)(t) &= int_0^th(t - v)cdot u(v) dv \ &= int_0^t e^{-(t-v)}(1-v)dv \ &= e^{-t} left( int_0^te^vdv - int_0^tve^vdv right) end{align}$$

Solving the first integral is rather easy:

$$int_0^te^vdv = left[e^vright]_0^t = e^t - 1$$

Solving the second integral requires integration by parts:

$$begin{align} int_0^tve^vdv &= int_0^tvd(e^v) \ &= left[vcdot e^vright]_0^t - int_0^tve^vdv end{align}$$

As you get the same integration as before, you can write:

$$begin{align} I &= left[ vcdot e^vright]_0^t - I \ Rightarrow 2I &= left[ vcdot e^vright]_0^t \ Rightarrow I &= frac{[vcdot e^v]_0^t}{2}=frac{te^t}{2} end{align}$$

Plugging it all in yields:

$$ begin{align} y(t) &= e^{-t} left( e^t - 1 - frac{te^t}{2} right) \ &= 1 - e^{-t} - frac{t}{2} end{align}$$

This answer is off by a factor 2, but it resembles the given answer. I don't see where I was wrong.

For $t geq 1$, you can write

$$ begin{align} y(t) &= int_0^1h(t-v)u(v)dv \ &= e^{-t} left( e - 1 - frac{e}{2} right) \ &= frac{e^{1-t}}{2} - e^{-t} end{align}$$

This is again off by the same factor of 2 compared to your reference.