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

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

Popular >

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

area

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

\frac{125}{6}

Decimal Notation

20.83333 \dots

Solution steps

Isolate

y

for

x = y : y = x

Isolate

y

for

x = - y^{2} + 6 y : y = 3 - - x + 9, y = - x + 9 + 3

f_{1} (x) = x

f_{2} (x) = 3 - - x + 9

To find the intersection points solve

f_{i} (x) = f_{j} (x)

x = 3 - - x + 9 : x = 0

x = - x + 9 + 3 : x = 5

3 - - x + 9 = - x + 9 + 3 : x = 9

= \int_{0}^{5} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{5}^{9} ∣ f_{2} (x) - f_{3} (x) ∣ d x

= \int_{0}^{5} x - (3 - - x + 9) d x + \int_{5}^{9} 3 - - x + 9 - (- x + 9 + 3) d x

\int_{0}^{5} x - (3 - - x + 9) d x = \frac{61}{6}

\int_{5}^{9} 3 - - x + 9 - (- x + 9 + 3) d x = \frac{32}{3}

= \frac{61}{6} + \frac{32}{3}

Simplify

= \frac{125}{6}

sin (x) + \frac{1}{7} cot (x)

\int_{50}^{80} \frac{25}{3} x d x

\int_{1}^{2} \frac{x ^{5} + 4 x ^{7}}{x ^{4}} d x

\frac{\partial}{\partial x} (cos (x yz))

f (x) = \frac{3 x + 2}{5 x - 1}