# Area enclosed by a curve and the x-axis?

Find the area enclosed between the curve y=x^3 - 4x^2 and the x-axis

### 3 Answers

- Brach ZLv 410 years agoFavourite answer
Find the area enclosed between the curve y=x^3 - 4x^2 and the x-axis.

f(x) = x³ - 4x² = x²(x - 4) intersects the x-axis where f(x) = 0, which is where x = 0 and 4.

The area between f(x) and the x-axis is bounded (finite) only when 0 ≤ x ≤ 4.

Elsewhere, the area is unbounded (infinite).

The area is the integral |⌠ (x³ - 4x²) dx| where 0 ≤ x ≤ 4

which is

|x^4 / 4 - 4/3 x^3| = |x³(x/4 - 4/3)| = | 64(-1/3) | = 64/3.

- Anonymous4 years ago
The confident imperative of ?(5x + 4) from x = 0 to x= a million can provide the asked area. imperative from ?(5x + 4) is (2/15)(5x +4)^(3/2) + c. Evaluated at a million and 0, that supplies us (2/15)*27 - (2/15)*8 = 38/15.

- henry_yang67Lv 610 years ago
y = x^2(x-4)

integrate 0-y for x from 0 to 4

-1/4 x^4 + 4/3 x^3 (x=4) = 256/12 = 64/3