Let

Put x – 4 = t³

⇒ dx = 3t²dt (Differentiating both sides)

When x = 4, t³ = 4 – 4 = 0 ⇒ t = 0

When x = 12, t³ = 12 – 4 = 8 ⇒ t = 2

So, the new limits are 0 and 2.

We can write x = t³ + 4

Substituting this in the original integral,

Recall,