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

The answer to area y=x,y=x^2,y=-x+1 is (6sqrt(5)-7)/6+3-sqrt(5)

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

area

y = x, y = x^{2}, y = - x + 1

\frac{6 5 - 7}{6} + 3 - 5

Decimal Notation

1.83333 \dots

Solution steps

To find the intersection points solve

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

x = x^{2} : x = 1, x = 0

x = - x + 1 : x = \frac{1}{2}

x^{2} = - x + 1 : x = \frac{- 1 + 5}{2}, x = \frac{- 1 - 5}{2}

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

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

\int_{\frac{- 1 - 5}{2}}^{0} x^{2} - (- x + 1) d x = \frac{7 + 5 5}{12}

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

\int_{\frac{1}{2}}^{\frac{- 1 + 5}{2}} ∣ x - (- x + 1) ∣ d x = - \frac{1}{4} + \frac{5 - 2 5}{2}

\int_{\frac{- 1 + 5}{2}}^{1} x^{2} - x d x = \frac{1}{2} - \frac{7 ( 3 - 5 )}{12}

= \frac{7 + 5 5}{12} + \frac{1}{4} - \frac{1}{4} + \frac{5 - 2 5}{2} + \frac{1}{2} - \frac{7 ( 3 - 5 )}{12}

Simplify

= \frac{6 5 - 7}{6} + 3 - 5

k = 1 \sum \infty k^{k} x^{k}

\frac{d}{d x} (\frac{2 x + 1}{2 x})

x^{2} - 2 x, - x^{2} + 4 x

\int \frac{2}{x + 2} d x

f (x) = 2 x + 1 + \frac{6}{x}, at x = 2

What is area y=x,y=x^2,y=-x+1 ?
The answer to area y=x,y=x^2,y=-x+1 is (6sqrt(5)-7)/6+3-sqrt(5)