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

The answer to area y=x^3-2x^2,y=2x^2-3x is 37/12

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

area

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

\frac{37}{12}

Decimal

3.08333 \dots

Solution steps

Apply the area formula:

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

Solve

\int_{0}^{3} x^{3} - 2 x^{2} - (2 x^{2} - 3 x) d x : \frac{37}{12}

The area is:

= \frac{37}{12}

in v erse l a pl a ce \frac{2}{( ( s + 2 ) ( s ^{2} - 4 ) )}

\int \frac{2}{- x ^{2} + 4 x} d x

\frac{d}{d θ} (4 + 4 sin (θ))

\frac{d}{d x} (x - 3 x^{\frac{1}{3}})

\frac{d}{d x} (cos (3 x) \cdot 3)

What is area y=x^3-2x^2,y=2x^2-3x ?
The answer to area y=x^3-2x^2,y=2x^2-3x is 37/12