What is area y=6x^2,y=4x^2,y=6-5x ?

The answer to area y=6x^2,y=4x^2,y=6-5x is 3191/864

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area y=6x^2,y=4x^2,y=6-5x

area

y = 6 x^{2}, y = 4 x^{2}, y = 6 - 5 x

\frac{3191}{864}

Decimal Notation

3.69328 \dots

Solution steps

To find the intersection points solve

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

6 x^{2} = 4 x^{2} : x = 0

6 x^{2} = 6 - 5 x : x = \frac{2}{3}, x = - \frac{3}{2}

4 x^{2} = 6 - 5 x : x = \frac{3}{4}, x = - 2

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

= \int_{- 2}^{- \frac{3}{2}} 4 x^{2} - (6 - 5 x) d x + \int_{- \frac{3}{2}}^{0} 6 x^{2} - 4 x^{2} d x + \int_{0}^{\frac{2}{3}} 6 x^{2} - 4 x^{2} d x + \int_{\frac{2}{3}}^{\frac{3}{4}} 4 x^{2} - (6 - 5 x) d x

\int_{- 2}^{- \frac{3}{2}} 4 x^{2} - (6 - 5 x) d x = \frac{29}{24}

\int_{- \frac{3}{2}}^{0} 6 x^{2} - 4 x^{2} d x = \frac{9}{4}

\int_{0}^{\frac{2}{3}} 6 x^{2} - 4 x^{2} d x = \frac{16}{81}

\int_{\frac{2}{3}}^{\frac{3}{4}} 4 x^{2} - (6 - 5 x) d x = \frac{97}{2592}

= \frac{29}{24} + \frac{9}{4} + \frac{16}{81} + \frac{97}{2592}

Simplify

= \frac{3191}{864}

\frac{d}{d x} (2 cos (x) + 2)

\frac{d}{d t} (e^{8 t})

\frac{d}{d x} (- \frac{1}{x ^{2}} - \frac{2}{x ^{3}})

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

f (x) = 2 x sin (3 x)

What is area y=6x^2,y=4x^2,y=6-5x ?
The answer to area y=6x^2,y=4x^2,y=6-5x is 3191/864