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

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

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

area

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

\frac{71}{6}

Solution steps

Apply the area formula:

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

Solve

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

The area is:

= \frac{71}{6}

\frac{d y}{d x} = x + 6 y

\frac{\partial}{\partial x} (cos (x + 5 y))

(2 x + 5) (- x + 2)

\int \frac{4 x ^{3} - x}{6 x ^{2} - 2 x - 1} d x

x \to 0 lim (\frac{( 3 x ^{2} - 2 x )}{x})

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