Finding enclosed area in calc- using integration?

Question

I need help in finding total area enclosed by the graphs of

y=5x^2−x^3+x and y=x^2+4x

Answer 1

Let y₁ = 5x² - x³ + x and y₂ = x² + 4x

y₁ - y₂ = 0⇒ (5x² - x³ + x) - (x² + 4x) = 0

4x² - x³ - 3x = 0 --> - x³ + 4x² - 3x = 0

Therefore  y₁ - y₂ = 0⇒ - x³ + 4x² - 3x = 0

Multiply each side by negative one.

y₂ - y₁ = 0 ⇒ x³ - 4x² + 3x = 0

x(x²  - 4x + 3) = 0

x = 0 and x²  - 4x + 3 = 0

x²  - 4x + 3 = 0

Now solve the equation using factor method.

x²  - 3x - 1x + 3 = 0.

x(x - 3) - 1(x - 3) = 0

Take out common factors.

(x - 3)(x - 1) = 0

x - 3 = 0 and x - 1 = 0

Simplify x = 3 and x = 1

There fore x = 0, 1, 3

The two equation interval is (0, 1, 3)

In interval 0<x<1, y₂>y₁ which makes first enclosed area A₁.

In the interval 1<x3, y₁>y₂ which is the second enclosed area A₂

y₂ - y₁  = x³ - 4x² + 3x

Cavalier's quadrate formula:

Rewrite the expression with common denominator.

There fore   

y₁ - y₂ = - x³ + 4x² - 3x

Rewrite the expression with common denominator.

Total area is A = A₁ + A₂