What is area y=3x,y= 3/5 x,y=28-x^2 ?

The answer to area y=3x,y= 3/5 x,y=28-x^2 is 8958/125

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area y=3x,y= 3/5 x,y=28-x^2

area

y = 3 x, y = \frac{3}{5} x, y = 28 - x^{2}

\frac{8958}{125}

Decimal Notation

71.664

Solution steps

To find the intersection points solve

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

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

3 x = 28 - x^{2} : x = - 7, x = 4

\frac{3}{5} x = 28 - x^{2} : x = - \frac{28}{5}, x = 5

= \int_{- 7}^{- \frac{28}{5}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{28}{5}}^{0} ∣ f_{2} (x) - f_{1} (x) ∣ d x + \int_{0}^{4} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{4}^{5} ∣ f_{2} (x) - f_{3} (x) ∣ d x

= \int_{- 7}^{- \frac{28}{5}} 3 x - (28 - x^{2}) d x + \int_{- \frac{28}{5}}^{0} \frac{3}{5} x - 3 x d x + \int_{0}^{4} 3 x - \frac{3}{5} x d x + \int_{4}^{5} \frac{3}{5} x - (28 - x^{2}) d x

\int_{- 7}^{- \frac{28}{5}} 3 x - (28 - x^{2}) d x = \frac{7399}{750}

\int_{- \frac{28}{5}}^{0} \frac{3}{5} x - 3 x d x = \frac{4704}{125}

\int_{0}^{4} 3 x - \frac{3}{5} x d x = \frac{96}{5}

\int_{4}^{5} \frac{3}{5} x - (28 - x^{2}) d x = \frac{149}{30}

= \frac{7399}{750} + \frac{4704}{125} + \frac{96}{5} + \frac{149}{30}

Simplify

= \frac{8958}{125}

\frac{\partial}{\partial x} (x^{5} e^{3 x y})

\frac{\partial}{\partial y} (2 + y e^{x y})

\int \frac{1}{4 + ( x - 2 ) ^{2}} d x

\frac{d y}{d t} = k y^{2}, y (0) = 47

\frac{\partial}{\partial x} ((x^{2} + y^{2})^{\frac{3}{2}})

What is area y=3x,y= 3/5 x,y=28-x^2 ?
The answer to area y=3x,y= 3/5 x,y=28-x^2 is 8958/125