Question

How do you do this question?

Answer 1

Step-by-step 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.