What is area x+y^2=2,x+y=0 ?

The answer to area x+y^2=2,x+y=0 is 9/2

area x+y^2=2,x+y=0

area

x + y^{2} = 2, x + y = 0

\frac{9}{2}

Decimal

4.5

Solution steps

Isolate

y

for

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

Isolate

y

for

x + y = 0 : y = - x

f_{1} (x) = 2 - x

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

To find the intersection points solve

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

2 - x = - 2 - x : x = 2

2 - x = - x : x = - 2

- 2 - x = - x : x = 1

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

= \int_{- 2}^{1} 2 - x - (- x) d x + \int_{1}^{2} - 2 - x - 2 - x d x

\int_{- 2}^{1} 2 - x - (- x) d x = \frac{19}{6}

\int_{1}^{2} - 2 - x - 2 - x d x = \frac{4}{3}

= \frac{19}{6} + \frac{4}{3}

Simplify

= \frac{9}{2}

\int_{0}^{3 π} 6 x^{2} sin (\frac{1}{6} x) d x

\int_{0}^{6} - x^{2} + 6 x d x

d er i v a t i v e 20 x^{3}

ma c l a u r in f (x) = cos (2 x)

d er i v a t i v e f (x) = (x^{3} - 4)^{2} (2 x + 5)^{3}