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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

area x,x^2,-x+1

area

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

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

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

\frac{d}{d x} (ln (\frac{x - 1}{x + 1}))

\frac{\partial}{\partial x} (\frac{m}{x ^{a} y ^{b} z ^{c}})

\frac{\partial}{\partial x} (\frac{3}{8} x y)

\int t^{2} e^{- (s - 1) t} d t

\frac{d}{d x} (a cos (b x + c) + d)

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