What is area 4sqrt(2x),2x^2,-4x+6 ?

The answer to area 4sqrt(2x),2x^2,-4x+6 is 67/3

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area 4sqrt(2x),2x^2,-4x+6

area

4 2 x, 2 x^{2}, - 4 x + 6

\frac{67}{3}

Decimal Notation

22.33333 \dots

Solution steps

To find the intersection points solve

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

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

4 2 x = - 4 x + 6 : x = \frac{1}{2}

2 x^{2} = - 4 x + 6 : x = 1, x = - 3

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

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

\int_{- 3}^{0} 2 x^{2} - (- 4 x + 6) d x = 18

\int_{0}^{\frac{1}{2}} 4 2 x - (- 4 x + 6) d x = \frac{7}{6}

\int_{\frac{1}{2}}^{1} 4 2 x - (- 4 x + 6) d x = - \frac{17}{6} + \frac{8 2}{3}

\int_{1}^{2} 2 x^{2} - 4 2 x d x = \frac{18 - 8 2}{3}

= 18 + \frac{7}{6} - \frac{17}{6} + \frac{8 2}{3} + \frac{18 - 8 2}{3}

Simplify

= \frac{67}{3}

(t + 2) y^{'} + y = \frac{1}{t - 1}

\frac{\partial}{\partial l} (2 π \frac{l}{g})

16 y^{^{''}} - π^{2} y = 0

\int x^{2} x^{3} + 15 d x

n = 1 \sum \infty e^{(- n^{2})}

What is area 4sqrt(2x),2x^2,-4x+6 ?
The answer to area 4sqrt(2x),2x^2,-4x+6 is 67/3