What is area y=4x,y= 7/5 x,y=32-x^2 ?

The answer to area y=4x,y= 7/5 x,y=32-x^2 is 70057/750

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area y=4x,y= 7/5 x,y=32-x^2

area

y = 4 x, y = \frac{7}{5} x, y = 32 - x^{2}

\frac{70057}{750}

Solution steps

To find the intersection points solve

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

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

4 x = 32 - x^{2} : x = - 8, x = 4

\frac{7}{5} x = 32 - x^{2} : x = - \frac{32}{5}, x = 5

= \int_{- 8}^{- \frac{32}{5}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{32}{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_{- 8}^{- \frac{32}{5}} 4 x - (32 - x^{2}) d x + \int_{- \frac{32}{5}}^{0} \frac{7}{5} x - 4 x d x + \int_{0}^{4} 4 x - \frac{7}{5} x d x + \int_{4}^{5} \frac{7}{5} x - (32 - x^{2}) d x

\int_{- 8}^{- \frac{32}{5}} 4 x - (32 - x^{2}) d x = \frac{5248}{375}

\int_{- \frac{32}{5}}^{0} \frac{7}{5} x - 4 x d x = \frac{6656}{125}

\int_{0}^{4} 4 x - \frac{7}{5} x d x = \frac{104}{5}

\int_{4}^{5} \frac{7}{5} x - (32 - x^{2}) d x = \frac{161}{30}

= \frac{5248}{375} + \frac{6656}{125} + \frac{104}{5} + \frac{161}{30}

Simplify

= \frac{70057}{750}

\frac{d}{d x} (\frac{x - 5 x x}{x})

\frac{d}{d x} (x e^{x} - 1)

f (x) = 5 x - 4 x^{2}, at x = - 1

\int_{0}^{3} (2^{x}) d x

y^{^{''}} + 2 y^{^{'}} + 4 y = 50 cos (x)

What is area y=4x,y= 7/5 x,y=32-x^2 ?
The answer to area y=4x,y= 7/5 x,y=32-x^2 is 70057/750