What is area 1+4x-x^2,11-x-x^2,1-(2x)/3 ?

The answer to area 1+4x-x^2,11-x-x^2,1-(2x)/3 is 725/18

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area 1+4x-x^2,11-x-x^2,1-(2x)/3

area

1 + 4 x - x^{2}, 11 - x - x^{2}, 1 - \frac{2 x}{3}

\frac{725}{18}

Decimal Notation

40.27777 \dots

Solution steps

To find the intersection points solve

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

1 + 4 x - x^{2} = 11 - x - x^{2} : x = 2

1 + 4 x - x^{2} = 1 - \frac{2 x}{3} : x = 0, x = \frac{14}{3}

11 - x - x^{2} = 1 - \frac{2 x}{3} : x = - \frac{10}{3}, x = 3

= \int_{- \frac{10}{3}}^{0} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{0}^{2} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{2}^{3} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{3}^{\frac{14}{3}} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- \frac{10}{3}}^{0} 11 - x - x^{2} - (1 - \frac{2 x}{3}) d x + \int_{0}^{2} 1 + 4 x - x^{2} - (11 - x - x^{2}) d x + \int_{2}^{3} 1 + 4 x - x^{2} - (11 - x - x^{2}) d x + \int_{3}^{\frac{14}{3}} 1 + 4 x - x^{2} - (1 - \frac{2 x}{3}) d x

\int_{- \frac{10}{3}}^{0} 11 - x - x^{2} - (1 - \frac{2 x}{3}) d x = \frac{1850}{81}

\int_{0}^{2} 1 + 4 x - x^{2} - (11 - x - x^{2}) d x = 10

\int_{2}^{3} 1 + 4 x - x^{2} - (11 - x - x^{2}) d x = \frac{5}{2}

\int_{3}^{\frac{14}{3}} 1 + 4 x - x^{2} - (1 - \frac{2 x}{3}) d x = \frac{400}{81}

= \frac{1850}{81} + 10 + \frac{5}{2} + \frac{400}{81}

Simplify

= \frac{725}{18}

x + 1, 2 x - 4, 0, 3

\int \frac{3 x ^{5} - 2}{4 x - 4 x ^{5}} d x

\frac{3}{x}, - 1

n = 1 \sum \infty cos (3 n)

\frac{d}{d x} (9 sin (5 x))

What is area 1+4x-x^2,11-x-x^2,1-(2x)/3 ?
The answer to area 1+4x-x^2,11-x-x^2,1-(2x)/3 is 725/18