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

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

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

area

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

\frac{3191}{864}

Decimal Notation

3.69328 \dots

Solution steps

Isolate

y

for

5 x + y = 6 : y = 6 - 5 x

f_{1} (x) = 4 x^{2}

f_{2} (x) = 6 x^{2}

To find the intersection points solve

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

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

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

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

= \int_{- 2}^{- \frac{3}{2}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{3}{2}}^{0} ∣ f_{2} (x) - f_{1} (x) ∣ d x + \int_{0}^{\frac{2}{3}} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{\frac{2}{3}}^{\frac{3}{4}} ∣ f_{1} (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}} 4 x^{2} - 6 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}} 4 x^{2} - 6 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}

(2 x - 3)^{2}

\frac{\partial}{\partial x} (x e^{- x} (A x^{2} + B x + C))

\int - \frac{ln ( x )}{x} d x

\frac{d y}{y - y ^{2}} = d x

\frac{d}{d x} (3 x y + sin (x) + z^{2})

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