What is area y=x^3-6x^2+8x,y=x^2-4x ?

The answer to area y=x^3-6x^2+8x,y=x^2-4x is 71/6

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area y=x^3-6x^2+8x,y=x^2-4x

area

y = x^{3} - 6 x^{2} + 8 x, y = x^{2} - 4 x

\frac{71}{6}

Decimal Notation

11.83333 \dots

Solution steps

Apply the area formula:

\int_{0}^{4} x^{3} - 6 x^{2} + 8 x - (x^{2} - 4 x) d x

Solve

\int_{0}^{4} x^{3} - 6 x^{2} + 8 x - (x^{2} - 4 x) d x : \frac{71}{6}

The area is:

= \frac{71}{6}

ln (x e^{x} + 2)

\frac{d}{d x} (\frac{2 cot ( x )}{csc ( x )})

\frac{d}{d x} (I n (- x^{2} + 6 x))

x \to 1 + lim (\frac{x + 9}{x + 2})

\frac{\partial}{\partial x} (4 y x^{3} - 12 y^{2} x^{2})

What is area y=x^3-6x^2+8x,y=x^2-4x ?
The answer to area y=x^3-6x^2+8x,y=x^2-4x is 71/6