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find the area enclosed?

+1 vote

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

asked Jan 31, 2013 in CALCULUS by anonymous Apprentice

1 Answer

+1 vote

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.

y - y = 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

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y₂ - y₁  = x3 - 12x2 + 35x

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Cavalieri's quadrature formula: \int x^a\,dx = \frac{x^{a+1}}{a+1} + C \qquad\text{(for } a\neq -1\text{)}\,\!

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y₁ - y₂ = - x3 + 12x2 - 35x

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A = - 444 + 4(218) - 420

A₂ = - 444 + 872 - 420 = 8

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

answered Jan 31, 2013 by richardson Scholar
edited Jan 31, 2013 by richardson

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