What is area 2y=4sqrt(x),y=4,2y+4x=8 ?

The answer to area 2y=4sqrt(x),y=4,2y+4x=8 is 11/3

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area 2y=4sqrt(x),y=4,2y+4x=8

area

2 y = 4 x, y = 4, 2 y + 4 x = 8

\frac{11}{3}

Decimal

3.66666 \dots

Solution steps

Isolate

y

for

2 y = 4 x : y = 2 x

Isolate

y

for

2 y + 4 x = 8 : y = 2 (- x + 2)

f_{1} (x) = 2 x

f_{2} (x) = 4

To find the intersection points solve

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

2 x = 4 : x = 4

2 x = 2 (- x + 2) : x = 1

4 = 2 (- x + 2) : x = 0

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

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

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

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

= 1 + \frac{8}{3}

Simplify

= \frac{11}{3}

\int \frac{d}{1} x^{2} d x

sin (3 x) + cos (8 x)

\frac{1}{3} x^{6} (x^{2} - 3)

\frac{d}{d x} ((5 - 2 x)^{- 2})

1 + 12 t

What is area 2y=4sqrt(x),y=4,2y+4x=8 ?
The answer to area 2y=4sqrt(x),y=4,2y+4x=8 is 11/3