Consider the following fictitious data:

    set.seed(2022)
    x1 = rnorm(30, 350, 50)
    x2 = rnorm(30, 300, 70)

Now do a two sample Welch t test of $H_0: \mu_1=\mu_2$
against $H_a: \mu_1 > \mu_2, using `t.test` in R:

    t.test(x1,x2, alt="gr")

            Welch Two Sample t-test

    data:  x1 and x2
    t = 2.6864, df = 55.074, p-value = 0.004764
    alternative hypothesis: 
     true difference in means is greater than 0
    95 percent confidence interval:
     13.8086     Inf
    sample estimates:
    mean of x mean of y 
     344.2034  307.5991 

The P-value of the test is computed by looking in the
upper tail of Student's t distribution with 55.074 degrees
of freedom. [DF is adjusted downward from $n_1+n_2-2=58$ to compensate for the difference
in sample variances.]

    1 - pt(2.6864, 55.074) 
    [1] 0.004764504

If you do a 2-sided t test, then the P-value is calculated
by looking in the lower tail below $-2.6864$ and above $2.6864.$  [By using `$`-notation we show only the P-value.]

    t.test(x1, x2)$p.val
    [1] 0.009528523

Computed as follows:

    pt(-2.6864, 55.074) + 1 - pt(2.6864, 55.074)  # left + right
    [1] 0.009529008

Alternatively, by the symmetry of the t distribution

    2*pt(-2.6864, 55.074)   $ Double left tail probability 
    [1] 0.009529008

Quantities in the output to the test are rounded slightly
to save space, so there is a tiny discrepancy with the P-values shown just above.

However, if you get confused (easy to do), and ask for the
wrong side, using parameter `alt="less"` in `t.test`, then
you get a nonsense P-value near $1.$

     t.test(x1, x2, alt="less")$p.val
     [1] 0.9952357