What is area y=-x^2+8x-12,y=x^2-8x+18,x=3x=6 ?

The answer to area y=-x^2+8x-12,y=x^2-8x+18,x=3x=6 is 8/3

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area y=-x^2+8x-12,y=x^2-8x+18,x=3x=6

area

y = - x^{2} + 8 x - 12, y = x^{2} - 8 x + 18, x = 3 x = 6

\frac{8}{3}

Decimal Notation

2.66666 \dots

Solution steps

Apply the area formula:

\int_{3}^{5} - x^{2} + 8 x - 12 - (x^{2} - 8 x + 18) d x

Solve

\int_{3}^{5} - x^{2} + 8 x - 12 - (x^{2} - 8 x + 18) d x : \frac{8}{3}

The area is:

= \frac{8}{3}

f (x) = 6 x^{2} + 4 x - 7, - 3

\int \frac{( arcsin ( \frac{x}{3} ) ) ^{2}}{9 - x ^{2}} d x

\frac{\partial}{\partial y} ((x - y)^{5})

\frac{d}{d x} (\frac{5 ^{x}}{ln ( 5 )})

\int (e - x - e^{2}) (e x + e - 5) d x

What is area y=-x^2+8x-12,y=x^2-8x+18,x=3x=6 ?
The answer to area y=-x^2+8x-12,y=x^2-8x+18,x=3x=6 is 8/3