Both dpois and dnorm in the code below sum to 1 (or thereabouts). This appears to confirm my understanding of the dpois and dnorm functions, which is that they represent proportions.
sum(dpois(x = 0:20, lambda = 5))
# 1
sum(dnorm(-10:30, mean = 10, sd = 2, log = FALSE))
# 1
But the output of dbeta does not sum to 1. Why is this?
sum(dbeta(c(0.1, 0.2, 0.3), shape1 = 2, shape2 = 4, ncp = 0, log = FALSE))
# 5.564
The relevant property of a probability density is not that it sums (for evaluation on some particular $x$ values) to one, but that it integrates to one.
If you evaluate a density $f$ at $x$ values that form a regular grid with grid width $\Delta x$, then you have very approximately
$$ \int_{-\infty}^\infty f(x)\,dx \approx \sum_{i=1}^n f(x_i)\Delta x, $$
which is why your initial two examples sum to almost one: here we have $\Delta x=1$, so we have approximately
$$ 1=\int_{-\infty}^\infty f(x)\,dx \approx \sum_{i=1}^n f(x_i). $$
Compare a finer grid for the normal case:
> sum(dnorm(seq(-10,30,by=0.1), mean = 10, sd = 2, log = FALSE))
[1] 10
And of course, if we include $\Delta x$ in your calculation for dbeta, then we again approximate the integral correctly:
> xx <- seq(0,1,by=0.01)
> sum(dbeta(xx, shape1 = 2, shape2 = 4, ncp = 0, log = FALSE)*mean(diff(xx)))
[1] 0.9998333
dpois is the probability mass function (pmf) a discrete Poisson distribution that can take integer values in $(0, \infty)$. If you sum over the probabilities for all integers, you should indeed get 1.
dnorm is the probability density function (pdf) for the normal distribution over the real numbers in $(\infty, \infty)$ and dbeta the pdf for a Beta distribution over $(0, 1)$. These will sum to 1, if you integrate them over their domain.
pdfs are the continuous distribution equivalent to pmfs, but that sum(dnorm(-10:30, mean = 10, sd = 2, log = FALSE)) sums to 1 is pure coincidence (or someone giving you a trick question).
sd had been 100, or if mean had been 13,712. Someone carefully chose the grid points and was careful to space them 1.0 apart to avoid the need for a scaling constant.
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sum(dnorm(0:20000, mean = 13712, sd = 100, log = FALSE)) gives 1 at least up to rounding (the range for the integers needs to cover the values where the density is of a noticeable size). Try sum(dnorm(-200:200,mean=0,sd=0.1,log=FALSE)) giving 3.989423 and sum(dnorm(-200:200,mean=0.5,sd=0.1,log=FALSE)) giving 0.0000297 for better counterexamples
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d* functions represent proportions only with a discrete response. In fact, your dnorm example just happens to sum to one, but
> sum(dnorm(seq(-10, 30, by = 0.01), mean = 10, sd = 2, log = FALSE))
[1] 100
is 100! The normal and beta distributions are continuous, not discrete. Therefore, instead of summing to 1, they must integrate to 1, which is different.
