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Finding enclosed area in calc- using integration?

+3 votes
I need help in finding total area enclosed by the graphs of

y=5x^2−x^3+x and y=x^2+4x
asked Feb 2, 2013 in CALCULUS by anonymous Apprentice

1 Answer

+3 votes
 
Best answer

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

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y₂ - y₁  = x3 - 4x2 + 3x

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

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Rewrite the expression with common denominator.

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There fore    image

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y₁ - y₂ = - x3 + 4x2 - 3x

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Rewrite the expression with common denominator.

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Total area is A = A₁ + A₂

 image

 

answered Feb 2, 2013 by richardson Scholar
selected Feb 3, 2013 by anonymous

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