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 Notation

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}

\frac{4}{( 11 x - 3 ) ^{9}}

\int_{0}^{\frac{π}{2}} cos (x) d x

\int \frac{x ^{2}}{100 + x ^{2}} d x

\int 5 (sec (x))^{2} d x

y = \frac{4}{3} (2 x - 1)^{3}

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