What is area 1-x^2,x^2-3,x+3 ?

The answer to area 1-x^2,x^2-3,x+3 is -2sqrt(2)+(59-20sqrt(2))/6+11

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area 1-x^2,x^2-3,x+3

area

1 - x^{2}, x^{2} - 3, x + 3

- 22 + \frac{59 - 20 2}{6} + 11

Decimal

13.29086 \dots

Solution steps

To find the intersection points solve

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

1 - x^{2} = x^{2} - 3 : x = 2, x = - 2

1 - x^{2} = x + 3 :

No Solution for

x \in R

x^{2} - 3 = x + 3 : x = 3, x = - 2

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

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

\int_{- 2}^{- 2} x^{2} - 3 - (x + 3) d x = - \frac{8 - 2 2}{3} + 11 - 62

\int_{- 2}^{2} 1 - x^{2} - (x + 3) d x = \frac{4 2}{3} + 42

\int_{2}^{3} x + 3 - (x^{2} - 3) d x = \frac{75 - 32 2}{6}

= - \frac{8 - 2 2}{3} + 11 - 62 + \frac{4 2}{3} + 42 + \frac{75 - 32 2}{6}

Simplify

= - 22 + \frac{59 - 20 2}{6} + 11

\frac{\partial}{\partial x} (2 (ln (x)) + y)

\frac{\partial}{\partial x} (x (x - 2))

\int \frac{1}{x ( x + 1 )} d x

ma c l a u r in 2 x e^{x^{2}}

\int x^{3} e^{x y}

What is area 1-x^2,x^2-3,x+3 ?
The answer to area 1-x^2,x^2-3,x+3 is -2sqrt(2)+(59-20sqrt(2))/6+11