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

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

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

area

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

\int - 10 cos (- 2 x + 7) d x

f (x) = sin (sin (x))

x^{2} d x = y^{3} d y

\frac{d}{d x} (lo g_{e} (3 x^{2} + 7))

\int sin^{3} (x) cos^{- 2} (x) d x

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