What is area f(x)=x^3-3x^2+3x,g(x)=x ?

The answer to area f(x)=x^3-3x^2+3x,g(x)=x is 1/2

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area f(x)=x^3-3x^2+3x,g(x)=x

area

f (x) = x^{3} - 3 x^{2} + 3 x, g (x) = x

\frac{1}{2}

Decimal Notation

0.5

Solution steps

Apply the area formula:

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

Solve

\int_{0}^{2} x^{3} - 3 x^{2} + 2 x d x : \frac{1}{2}

The area is:

= \frac{1}{2}

f (3) = 2 x^{2} + 4 x

\int \frac{1}{x ^{2} 25 x ^{2} + 64} d x

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

f (x) = x^{2} + 4 x, g (x) = x + 10

\frac{3}{s + 5}

What is area f(x)=x^3-3x^2+3x,g(x)=x ?
The answer to area f(x)=x^3-3x^2+3x,g(x)=x is 1/2