A continuous random variable $X$ has the density $f_X$:
$$ f_x(x) = \begin{cases}\frac{x}{a}, & 20 < x < 40,\\0, & \textrm{otherwise.}\end{cases} $$
Find $a$ so that $f_x(x)$ becomes a proper density function.
I'm aware that $f$'s total area under the curve has to be $1$, which is the core of the problem. E.g:
$$ \int_{-\infty}^\infty f_xdx = \int_{21}^{39} f_xdx = 1 $$
Solving this yields $a = 540$. However, my learning material's solution has different integral borders:
$$ \int_{20}^{40} f_xdx = 1 $$
Which gives another answer. Is this simply a typo, considering $f_x(20) = f_x(40) = 0$. Or why should $20$ and $40$ be included in the evaluation?
Remember that when we calculate the area under a curve, excluding only point does not affect the value of the area. So you can consider this as as the function
$f(x)= \frac{x}{a}$ where $20\le x\le 40$
Why is that? Go back to the definition of integral, it's the limit of Riemann sum, Will Riemann sum be affected by just excluding a point? absolutely not! this can be seen as intuitively true.