What is area 4x+y^2=12,x=y ?

The answer to area 4x+y^2=12,x=y is 64/3

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area 4x+y^2=12,x=y

area

4 x + y^{2} = 12, x = y

\frac{64}{3}

Decimal Notation

21.33333 \dots

Solution steps

Isolate

y

for

4 x + y^{2} = 12 : y = 2 3 - x, y = - 2 3 - x

Isolate

y

for

x = y : y = x

f_{1} (x) = 2 3 - x

f_{2} (x) = - 2 3 - x

To find the intersection points solve

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

2 3 - x = - 2 3 - x : x = 3

2 3 - x = x : x = 2

- 2 3 - x = x : x = - 6

= \int_{- 6}^{2} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{2}^{3} ∣ f_{1} (x) - f_{2} (x) ∣ d x

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

\int_{- 6}^{2} - 2 3 - x - x d x = \frac{56}{3}

\int_{2}^{3} 2 3 - x - (- 2 3 - x) d x = \frac{8}{3}

= \frac{56}{3} + \frac{8}{3}

Simplify

= \frac{64}{3}

\int \frac{x ^{4} + x ^{2} - 1}{x - 1} d x

\frac{( s + 2 )}{s ^{2} + 2 s + 2}

x \to a lim (\frac{ln ( x ) - ln ( a )}{x - a})

\int (\frac{sec ( x )}{tan ( x )})^{4} d x

\frac{d y}{d x} = e^{4 x - 3 y}, y (0) = 2