What is area f(x)=x^3-4x^2+4x,[0,3] ?

The answer to area f(x)=x^3-4x^2+4x,[0,3] is 9/4

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area f(x)=x^3-4x^2+4x,[0,3]

area

f (x) = x^{3} - 4 x^{2} + 4 x, [0, 3]

\frac{9}{4}

Solution steps

Apply the area formula:

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

Solve

\int_{0}^{3} x^{3} - 4 x^{2} + 4 x d x : \frac{9}{4}

The area is:

= \frac{9}{4}

\frac{3 x}{x ^{2} + 3}

y^{^{'}} = \frac{x}{3 y}

\int \frac{6}{1 - cos ( 3 x )} d x

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

\int x 1 0^{x} d x

What is area f(x)=x^3-4x^2+4x,[0,3] ?
The answer to area f(x)=x^3-4x^2+4x,[0,3] is 9/4