Answer:
Explanation:
Let A be the area of the surface of the cube.
When the side length changes from s to 4s, the new area A' can be calculated as:
A' = 6(4s)^2 = 96s^2
The change in area is then:
ΔA = A' - A = 96s^2 - 6s^2 = 90s^2
To find the integral that quantifies the change in area, we can integrate the expression for ΔA with respect to s, from s to 4s:
∫(90s^2)ds from s to 4s
= [30s^3] from s to 4s
= 30(4s)^3 - 30s^3
= 1920s^3 - 30s^3
= 1890s^3
Therefore, the integral that quantifies the change in area of the surface of a cube when its side length quadruples from s units to 4s units is:
∫(90s^2)ds from s to 4s
= 1890s^3 from s to 4s
= 1890(4s)^3 - 1890s^3
= 477,840s^3 - 1890s^3