What is area y=sqrt(2x),2sqrt(x-4),y=0 ?

The answer to area y=sqrt(2x),2sqrt(x-4),y=0 is 32/3

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area y=sqrt(2x),2sqrt(x-4),y=0

area

y = 2 x, 2 x - 4, y = 0

\frac{32}{3}

Decimal

10.66666 \dots

Solution steps

To find the intersection points solve

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

2 x = 0 : x = 0

2 x = 2 x - 4 : x = 8

0 = 2 x - 4 : x = 4

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

= \int_{0}^{4} 2 x - 0 d x + \int_{4}^{8} 2 x - 2 x - 4 d x

\int_{0}^{4} 2 x - 0 d x = - \frac{8 2}{3} + 82

\int_{4}^{8} 2 x - 2 x - 4 d x = \frac{32 + 8 2}{3} - 82

= - \frac{8 2}{3} + 82 + \frac{32 + 8 2}{3} - 82

Simplify

= \frac{32}{3}

5 x^{3} - 10 x^{2} - \frac{3}{x}

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

\frac{d y}{d t} = 5 y + y^{4}, y (0) = 1

1 - cos^{2} (17 x)

f (x) = 6 x^{2} - 20 x^{3}

What is area y=sqrt(2x),2sqrt(x-4),y=0 ?
The answer to area y=sqrt(2x),2sqrt(x-4),y=0 is 32/3