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

The answer to area y=-x^2+6x,y=x^2-2x is 64/3

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area y=-x^2+6x,y=x^2-2x

area

y = - x^{2} + 6 x, y = x^{2} - 2 x

\frac{64}{3}

Decimal Notation

21.33333 \dots

Solution steps

Apply the area formula:

\int_{0}^{4} - x^{2} + 6 x - (x^{2} - 2 x) d x

Solve

\int_{0}^{4} - x^{2} + 6 x - (x^{2} - 2 x) d x : \frac{64}{3}

The area is:

= \frac{64}{3}

\frac{3}{5 x ^{4}}

4 y^{^{''}} + 3 y^{^{'}} + 7 y = 0

x y^{^{'}} + y = x^{2} + 1

\frac{d}{d x} ((3 cos (2 x) + sin (3 x))^{2})

\int (x^{- 1}) d x