What is area 5x, 1/7 x,50-x^2 ?

The answer to area 5x, 1/7 x,50-x^2 is 271337/1029

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area 5x, 1/7 x,50-x^2

area

5 x, \frac{1}{7} x, 50 - x^{2}

\frac{271337}{1029}

Decimal Notation

263.68999 \dots

Solution steps

To find the intersection points solve

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

5 x = \frac{1}{7} x : x = 0

5 x = 50 - x^{2} : x = - 10, x = 5

\frac{1}{7} x = 50 - x^{2} : x = - \frac{50}{7}, x = 7

= \int_{- 10}^{- \frac{50}{7}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{50}{7}}^{0} ∣ f_{2} (x) - f_{1} (x) ∣ d x + \int_{0}^{5} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{5}^{7} ∣ f_{2} (x) - f_{3} (x) ∣ d x

= \int_{- 10}^{- \frac{50}{7}} 5 x - (50 - x^{2}) d x + \int_{- \frac{50}{7}}^{0} \frac{1}{7} x - 5 x d x + \int_{0}^{5} 5 x - \frac{1}{7} x d x + \int_{5}^{7} \frac{1}{7} x - (50 - x^{2}) d x

\int_{- 10}^{- \frac{50}{7}} 5 x - (50 - x^{2}) d x = \frac{55000}{1029}

\int_{- \frac{50}{7}}^{0} \frac{1}{7} x - 5 x d x = \frac{42500}{343}

\int_{0}^{5} 5 x - \frac{1}{7} x d x = \frac{425}{7}

\int_{5}^{7} \frac{1}{7} x - (50 - x^{2}) d x = \frac{538}{21}

= \frac{55000}{1029} + \frac{42500}{343} + \frac{425}{7} + \frac{538}{21}

Simplify

= \frac{271337}{1029}

(x^{2} + 1) y^{^{'}} + 7 x (y - 1) = 0, y (0) = 4

x \to \infty lim (cos (0))

sin (x) + \frac{5}{8} cot (x)

x \to - 6 lim (x^{2} - 1)

n = 1 \sum \infty \frac{1}{e ^{2 n}}

What is area 5x, 1/7 x,50-x^2 ?
The answer to area 5x, 1/7 x,50-x^2 is 271337/1029