What is area y=1-x^2,y=2x,y= 1/2 x ?

The answer to area y=1-x^2,y=2x,y= 1/2 x is -2sqrt(2)+(32sqrt(2)-35)/(16)+5

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area y=1-x^2,y=2x,y= 1/2 x

area

y = 1 - x^{2}, y = 2 x, y = \frac{1}{2} x

- 22 + \frac{32 2 - 35}{16} + 5

Decimal Notation

2.8125

Solution steps

To find the intersection points solve

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

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

1 - x^{2} = \frac{1}{2} x : x = - \frac{1 + 17}{4}, x = \frac{17 - 1}{4}

2 x = \frac{1}{2} x : x = 0

= \int_{- 1 - 2}^{- \frac{1 + 17}{4}} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{- \frac{1 + 17}{4}}^{0} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{0}^{2 - 1} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{2 - 1}^{\frac{17 - 1}{4}} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 1 - 2}^{- \frac{1 + 17}{4}} 1 - x^{2} - 2 x d x + \int_{- \frac{1 + 17}{4}}^{0} 2 x - \frac{1}{2} x d x + \int_{0}^{2 - 1} 2 x - \frac{1}{2} x d x + \int_{2 - 1}^{\frac{17 - 1}{4}} 1 - x^{2} - \frac{1}{2} x d x

\int_{- 1 - 2}^{- \frac{1 + 17}{4}} 1 - x^{2} - 2 x d x = 2 + \frac{16 2 - 21 - 13 17}{48} + 1

\int_{- \frac{1 + 17}{4}}^{0} 2 x - \frac{1}{2} x d x = \frac{3 ( 9 + 17 )}{32}

\int_{0}^{2 - 1} 2 x - \frac{1}{2} x d x = 3 - 22 - \frac{3 - 2 2}{4}

\int_{2 - 1}^{\frac{17 - 1}{4}} 1 - x^{2} - \frac{1}{2} x d x = - 2 + \frac{17 17 + 112 2 - 177}{96} + 1

= 2 + \frac{16 2 - 21 - 13 17}{48} + 1 + \frac{3 ( 9 + 17 )}{32} + 3 - 22 - \frac{3 - 2 2}{4} - 2 + \frac{17 17 + 112 2 - 177}{96} + 1

Simplify

= - 22 + \frac{32 2 - 35}{16} + 5

\int_{0}^{1} e^{3 x} - e^{x} d x

\int \frac{y + 1}{y ^{2} + 2 y - 4} d y

\int 9 x^{4} d x

x \to 3 - lim (c x^{2} + 2 x)

x^{^{'}} = \frac{1}{x ^{2}}

What is area y=1-x^2,y=2x,y= 1/2 x ?
The answer to area y=1-x^2,y=2x,y= 1/2 x is -2sqrt(2)+(32sqrt(2)-35)/(16)+5