$$
h(u) = \int_{-\infty}^\infty f(x) g(u-x)\,dx.
$$
In the expression $\displaystyle \int_{-\infty}^\infty \cdots\cdots\,dx,$ anything that remains fixed as $x$ goes from $-\infty$ to $+\infty$ is a “constant.” Thus $\dfrac d{dx} (u-x) = 0-1,\vphantom{\dfrac{\displaystyle\sum}{}}$ so if we set $y=u-x,$ then $dy = -dx.$
Fourier coefficients are complex numbers, so you expect to get complex numbers. (It is perhaps helpful to think of these numbers as encoding the magnitude (at all times) and phase (at time $0$) of the corresponding frequency components.
We use Euler's formula to rewrite the exponential in your integral in terms of sine and cosine.
$$ \mathrm{e}^{\mathrm{i}\theta} = \cos \theta + \mathrm{i}\,\sin \theta \text{,} $$
so (using even-odd properties in the last line)
\begin{align*}
\mathrm{e}^{\mathrm{i} (-nt)}
&= \cos(-nt) + \mathrm{i}\,\sin(-nt) \\
&= \cos(nt) - \mathrm{i} \,\sin(nt) \text{.}
\end{align*}
Now let's rewrite your function to track the real and imaginary parts in separate integrals, turning both integrals into more familiar real integrals.
\begin{align*}
\frac{1}{2\pi} &\int_0^{2\pi} \; f(t) \mathrm{e}^{\mathrm{i} (-nt)} \,\mathrm{d}t \\
&= \frac{1}{2\pi} \int_0^{2\pi} \; f(t) (\cos(nt) - \mathrm{i} \,\sin(nt)) \,\mathrm{d}t \\
&= \frac{1}{2\pi} \int_0^{2\pi} \; f(t) \cos(nt) \,\mathrm{d}t - \frac{\mathrm{i}}{2\pi} \int_0^{2\pi} \; f(t)\sin(nt)) \,\mathrm{d}t \text{.}
\end{align*}
These are the Fourier sine and cosine transforms. These two integrands are real valued everywhere, so your thought of partitioning and summing areas of rectangles works. For each $n$ you will get a complex coefficient $u_n + \mathrm{i}\, v_n$. To get the amplitude and phase, convert to the polar representation, $r_n \mathrm{e}^{\mathrm{i}\,\theta_n} = u_n + \mathrm{i}\, v_n$.
Let $g = \Re f$ and $h = \Im f$ so that $g$ and $h$ are both real functions and $f(t) = g(t) + \mathrm{i}\, h(t)$. Then
\begin{align*}
\frac{1}{2\pi} &\int_0^{2\pi} f(t) \mathrm{e}^{\mathrm{i}(-nt)} \,\mathrm{d}t \\
&= \frac{1}{2\pi} \int_0^{2\pi} (g(t) + \mathrm{i}\, h(t)) (\cos(nt) - \mathrm{i} \,\sin(nt)) \,\mathrm{d}t \\
&= \frac{1}{2\pi}\left(
\int_0^{2\pi} g(t)\cos(nt) \,\mathrm{d}t
- \mathrm{i}\int_0^{2\pi} g(t)\sin(nt) \,\mathrm{d}t
+ \int_0^{2\pi} h(t)\sin(nt)\,\mathrm{d}t
+ \mathrm{i}\int_0^{2\pi} h(t)\cos(nt) \,\mathrm{d}t
\right) \text{,}
\end{align*}
and each of these four integrands is a real function so your model of finding area under a curve applies to each.
Best Answer
We are asked to graphically find the convolution of the two pulses:
We see that for $f(t), t \in [0,5]$ and for $g(t), t \in [0, 1]$, thus we need to evaluate the convolution integral in the range $[0 + 0, 5 + 1] = [0, 6]$, that is $0 \le t \le 6$.
We need to flip one of the pulses about the vertical axis and since they are pulses, we can choose either one, but it usually easier to choose the one with smaller duration, so lets choose to flip $g(t)$. We also now write these as $f(\tau), g(\tau)$. A plot of these is:
If we shift $g(- \tau)$ to the left, we get $g(t-\tau)$, which we can see in the plot above, we have no overlap between the two curves, thus the integral is zero in this region (which we will call it region $1$). Note that we label $g(-\tau)$ over the range $(t-1, t)$ for all calculations (notice what you get for $t=1$).
Next, we start shifting the signal $g(t-\tau)$ to the right $(t \gt 0)$. Consider first the interval $0 \le t \le 1$, which is:
In this interval, the signals overlap as shown, hence the product is not zero in this interval and the convolution is given as (which we call region $2$):
$$\displaystyle f(t) \ast g(t) = \int_0^t 1 \times 2 ~d\tau = 2t, 0 \le t \le 1$$
By shifting $g(t-\tau)$ further to the right, we get the same kind of overlap for $1 \le t \le 5$ as:
In this interval, which we will call region $3$, the integral yields:
$$ f(t) \ast g(t) = \displaystyle \int_{t-1}^t 1 \times 2 ~d\tau = 2, 1 \le t \le 5$$
By shifting $g(t-\tau)$ further to the right , for $5 \le t \le 6$, we get:
Again, we have overlap (which we will call region $4$), so the integral is given by:
$$ f(t) \ast g(t) = \displaystyle \int_{t-1}^5 1 \times 2 ~d\tau = 2(6-t), 5 \le t \le 6$$
Lastly, we shift $g(t-\tau)$ further to the right for $t \ge 6$ and we have no overlap (we will call this region $5$), so the integral is zero, as:
Putting all of these regions together, we have:
$$f(t) \ast g(t) = \begin{cases}\begin{array}{cc} 0 & t\leq 0 \\ 2 t & 0\leq t\leq 1 \\ 2 & 1\leq t\leq 5 \\ 2 (6-t) & 5\leq t\leq 6 \\ 0 & t \ge 6 \end{array} \end{cases} $$
A plot of the convolution $f(t) \ast g(t)$ is:
For the second part, we can write the functions as:
$$f(t) = u(t) - u(t-5), g(t) = 2(u(t) - u(t-1))$$
Now, we need to find:
$$f(t) \ast g(t) = \displaystyle \int_{- \infty}^{\infty} f(\tau) \ast g(t-\tau)~ d\tau = \int_0^6 f(\tau) \ast g(t-\tau)~ d\tau$$
We can also solve this as:
$$f(t) \ast g(t) = \int_{- \infty}^{\infty} g(\tau) \ast f(t-\tau)~ d\tau = \int_0^6 g(\tau) \ast f(t-\tau)~ d\tau $$
I will let you work out those details as you already know the answer.