What is area 4x=4y-y^2,4x-y=0 ?

The answer to area 4x=4y-y^2,4x-y=0 is 9/8

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

area

4 x = 4 y - y^{2}, 4 x - y = 0

\frac{9}{8}

Decimal Notation

1.125

Solution steps

Isolate

y

for

4 x = 4 y - y^{2} : y = - 2 (- x + 1 - 1), y = 2 (- x + 1 + 1)

Isolate

y

for

4 x - y = 0 : y = 4 x

f_{1} (x) = - 2 (- x + 1 - 1)

f_{2} (x) = 2 (- x + 1 + 1)

To find the intersection points solve

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

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

- 2 (- x + 1 - 1) = 4 x : x = 0

2 (- x + 1 + 1) = 4 x : x = \frac{3}{4}

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

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

\int_{0}^{\frac{3}{4}} - 2 (- x + 1 - 1) - 4 x d x = \frac{19}{24}

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

= \frac{19}{24} + \frac{1}{3}

Simplify

= \frac{9}{8}

\int \frac{1}{x ^{2} 7 + x ^{2}} d x

f (x) = e^{7 x} cos (10 x)

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

\int_{- 1}^{1} x^{2} (1.5 x^{2}) d x

a \cdot t (8 - a) + 8 a