find the area enclosed?

Question

y=13x^2-x^3+x and y=x^2+36x

Answer 1

Let y₁ = 13x² - x³ + x and y₂ = x² + 36x

y₁ - y₂ = 0⇒ (13x² - x³ + x) - (x² + 36x) = 0

13x² - x³ + x - x² - 36x = 0

12x² - x³ - 35x = 0

- x³ + 12x² - 35x = 0

There fore  y₁ - y₂ = 0⇒ - x³ + 12x² - 35x = 0

Multiply each side by negative one.

y₂ - y₁ = 0 ⇒ x³ - 12x² + 35x = 0

 x³  - 12x² + 35x  = 0

x(x²  - 12x + 35) = 0

x = 0 and x²  - 12x + 35 = 0

x²  - 12x + 35 = 0

Now slove the factor method.

x²  - 7x - 5x + 35 = 0.

x(x - 7) - 5(x - 7) = 0

Take out common factors.

(x - 7)(x - 5) = 0

x - 7 = 0 and x - 5 = 0

There fore x = 0, 5, 7

The two equation interval is (0, 5, 7)

In interval 0<x<5, y₂>y₁ which makes first enclosed area A₁.

In the interval 5<x<7, y₁>y₂ which is the second enclosed area A₂

y₂ - y₁  = x³ - 12x² + 35x

Cavalieri's quadrature formula:

 

y₁ - y₂ = - x³ + 12x² - 35x

A₂ = - 444 + 4(218) - 420

A₂ = - 444 + 872 - 420 = 8

Total area is A = A1 + A2 = 93.75 + 8 = 101.75 squre unit