Question

Find the area of the region between y=3x^(2) and y=x^(2)+4.

Answer 1

(16/3)√2 ≈ 7.542

Explanation:

You want the area between the curves y = 3x² and y = x² +4.

Area

The area will be the integral over the region where the second curve exceeds the first curve. The boundaries of that region are ...

3x² = x² +4

2x² = 4

x² = 2

x = ±√2

Integral

Each curve is symmetrical about the y-axis, so we can double the integral over half the region:

`\displaystyle A=2\int_0^(√(2)){((x^2+4)-3x^2)}\,dx=2\left[4x-(2x^3)/(3)-\right]_0^(√(2))\\\\\\A=2√(2)\left(4-(4)/(3)\right)=\boxed{(16√(2))/(3)}`

__

Additional comment

The area inside a parabola is 2/3 of the area of the bounding rectangle (for symmetrical portions including the vertex). This means the area will be ...

A = 2/3(2√2)(6 -2) = (16/3)√2

<95141404393>