What is area y=x^3-6x^2+9x,y=x,x=0,x=2 ?

The answer to area y=x^3-6x^2+9x,y=x,x=0,x=2 is 4

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area y=x^3-6x^2+9x,y=x,x=0,x=2

area

y = x^{3} - 6 x^{2} + 9 x, y = x, x = 0, x = 2

4

Solution steps

Apply the area formula:

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

Solve

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

The area is:

= 4

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\frac{d}{d x} (x^{2} sin (x) + x)

f (x) = \frac{2}{x + 1}

\frac{( x ^{4} - 5 x ^{3} + x ^{\frac{1}{2}} )}{x ^{2}}

\int_{0}^{\infty} 4 x e^{- 2 x} d x

What is area y=x^3-6x^2+9x,y=x,x=0,x=2 ?
The answer to area y=x^3-6x^2+9x,y=x,x=0,x=2 is 4