What is area x=-y^2+6,y= x/2-3/2 ?

The answer to area x=-y^2+6,y= x/2-3/2 is 32/3

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area x=-y^2+6,y= x/2-3/2

area

x = - y^{2} + 6, y = \frac{x}{2} - \frac{3}{2}

\frac{32}{3}

Decimal

10.66666 \dots

Solution steps

Isolate

y

for

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

f_{1} (x) = - x + 6

f_{2} (x) = - - x + 6

To find the intersection points solve

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

- x + 6 = - - x + 6 : x = 6

- x + 6 = \frac{x}{2} - \frac{3}{2} : x = 5

- - x + 6 = \frac{x}{2} - \frac{3}{2} : x = - 3

= \int_{- 3}^{5} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{5}^{6} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{- 3}^{5} - - x + 6 - (\frac{x}{2} - \frac{3}{2}) d x + \int_{5}^{6} - x + 6 - (- - x + 6) d x

\int_{- 3}^{5} - - x + 6 - (\frac{x}{2} - \frac{3}{2}) d x = \frac{28}{3}

\int_{5}^{6} - x + 6 - (- - x + 6) d x = \frac{4}{3}

= \frac{28}{3} + \frac{4}{3}

Simplify

= \frac{32}{3}

What is area x=-y^2+6,y= x/2-3/2 ?
The answer to area x=-y^2+6,y= x/2-3/2 is 32/3