Equivalence of the dt and pt function in R in my own curiosity I decided to question the equivalence of dt and pt in R.
pt denotes the cumulative density function of the t-distribution. So this gives us $P(X \le x)$. So if I want to find the probability of $P(X > x)$ I need to do $1 - P(X \le x)$. So far so good. You might do this if you are calculating p-values for a t-test. 
But lets say I want to use the regular probability function dt. I believe this calculates the probability of $P(X > x)$ occurring (since $P(X = x) = 0$ in a continuous distribution). I would expect $pt$ and $1-dt$ to be the same.
However in R I get -
> 1-dt(3, df=10)
[1] 0.9885995
> pt(3, df=10)
[1] 0.9933282

They are close but not equal. They are far enough away though that I would think this is some sort of error in my logic, and not some sort of error with their numerical solution.
Can anyone fill in the gaps here? Thank you!
 A: 
I believe this calculates the probability of P(X>x) occurring (since P(X=x)=0 in a continuous distribution). I would expect pt and 1−dt to be the same.

Your belief (and the consequently the expectation you hold) is wrong.
The $d$ in dt refers to density. You're right to think density is not probability. The d... functions in R (when done on continuous distributions like the $t$) don't return probability, they return the height of the density function at the value of their first argument.
https://en.wikipedia.org/wiki/Probability_density_function
If you look at the wikipedia page on the t-distribution, the top diagram on the right shows the density -- the thing returned by dt for several different degrees of freedom. The diagram below it shows the distribution function (cdf), which is the thing returned by pt (by default at least). If you want the upper tail, you can get that by changing the arguments to pt. This is explained in the help on the t-functions ?dt
pt(x,df) returns the area under the density to the left of x:

