What is area y=3x,y= 5/7 x,y=54-x^2 ?

The answer to area y=3x,y= 5/7 x,y=54-x^2 is 131576/1029

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area y=3x,y= 5/7 x,y=54-x^2

area

y = 3 x, y = \frac{5}{7} x, y = 54 - x^{2}

\frac{131576}{1029}

Decimal Notation

127.86783 \dots

Solution steps

To find the intersection points solve

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

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

3 x = 54 - x^{2} : x = - 9, x = 6

\frac{5}{7} x = 54 - x^{2} : x = - \frac{54}{7}, x = 7

= \int_{- 9}^{- \frac{54}{7}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{54}{7}}^{0} ∣ f_{2} (x) - f_{1} (x) ∣ d x + \int_{0}^{6} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{6}^{7} ∣ f_{2} (x) - f_{3} (x) ∣ d x

= \int_{- 9}^{- \frac{54}{7}} 3 x - (54 - x^{2}) d x + \int_{- \frac{54}{7}}^{0} \frac{5}{7} x - 3 x d x + \int_{0}^{6} 3 x - \frac{5}{7} x d x + \int_{6}^{7} \frac{5}{7} x - (54 - x^{2}) d x

\int_{- 9}^{- \frac{54}{7}} 3 x - (54 - x^{2}) d x = \frac{8019}{686}

\int_{- \frac{54}{7}}^{0} \frac{5}{7} x - 3 x d x = \frac{23328}{343}

\int_{0}^{6} 3 x - \frac{5}{7} x d x = \frac{288}{7}

\int_{6}^{7} \frac{5}{7} x - (54 - x^{2}) d x = \frac{295}{42}

= \frac{8019}{686} + \frac{23328}{343} + \frac{288}{7} + \frac{295}{42}

Simplify

= \frac{131576}{1029}

\int arcsin (\frac{1}{x}) d x

f (x) = 8.8 x

\frac{d}{d x} (e^{2 x} x (a cos (x) + b sin (x)))

y^{^{''}} + 3 y^{^{'}} - 4 y = - 8 e^{x} cos (2 x)

\frac{d}{d x} (\frac{1}{2} x - \frac{1}{10})

What is area y=3x,y= 5/7 x,y=54-x^2 ?
The answer to area y=3x,y= 5/7 x,y=54-x^2 is 131576/1029