Explanation:
f(x) = 33 / √(3 − x) has a vertical asymptote at x = 3. So we can change this integral to:
lim(t→3) ∫₂ᵗ 33 / √(3 − x) dx
If u = 3 − x, then du = -dx.
∫ -33 / √u du
-66 √u
Substitute back:
-66 √(3 − x)
Evaluate between x=2 and x=t.
[-66 √(3 − t)] − [-66 √(3 − 2)]
-66 √(3 − t) + 66
Plug into the limit:
lim(t→3) [-66 √(3 − t) + 66]
66
The limit exists, so the integral converges.