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

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

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

area

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

\frac{64}{3}

Solution steps

Apply the area formula:

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

Solve

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

The area is:

= \frac{64}{3}

x \to 3 lim (- (\frac{2}{x - 3}))

f (x) = 7 e^{(3 x^{2} - 2 x + 7)}

y^{^{'}} = \frac{1}{sec ^{2} ( x )}

\frac{d}{d x} (x^{2} + y^{2} + z^{2})

y = cos (x), y = 2 - cos (x), 0 \leq x \leq 2 π