What is area y=x^2-6x,y=-x^2+4x ?

The answer to area y=x^2-6x,y=-x^2+4x is 125/3

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area y=x^2-6x,y=-x^2+4x

area

y = x^{2} - 6 x, y = - x^{2} + 4 x

\frac{125}{3}

Decimal

41.66666 \dots

Solution steps

Apply the area formula:

\int_{0}^{5} x^{2} - 6 x - (- x^{2} + 4 x) d x

Solve

\int_{0}^{5} x^{2} - 6 x - (- x^{2} + 4 x) d x : \frac{125}{3}

The area is:

= \frac{125}{3}

y^{^{'}} = - 50 k y + k

cos (θ)

\int_{0}^{2} - t^{3} d t

\int \frac{2 x ^{3} + 3 x}{( x ^{4} + 3 x ^{2} + 7 ) ^{4}} d x

\int_{1}^{2} (1 + 2 y)^{2} d y

What is area y=x^2-6x,y=-x^2+4x ?
The answer to area y=x^2-6x,y=-x^2+4x is 125/3