What is area 2y=x,(y^2)/2 =x ?

The answer to area 2y=x,(y^2)/2 =x is 16/3

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area 2y=x,(y^2)/2 =x

area

2 y = x, \frac{y ^{2}}{2} = x

\frac{16}{3}

Decimal Notation

5.33333 \dots

Solution steps

Isolate

y

for

2 y = x : y = \frac{x}{2}

Isolate

y

for

\frac{y ^{2}}{2} = x : y = 2 x, y = - 2 x

f_{1} (x) = \frac{x}{2}

f_{2} (x) = 2 x

To find the intersection points solve

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

\frac{x}{2} = 2 x : x = 8, x = 0

\frac{x}{2} = - 2 x : x = 0

2 x = - 2 x : x = 0

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

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

\int_{0}^{8} \frac{x}{2} - 2 x d x = \frac{16}{3}

= \frac{16}{3}

\frac{d}{d x} (arcsec (x + 2))

x \to \infty lim (\frac{1}{- x})

\int (2 x^{3} - \frac{2}{3} x^{2} + 5 x) d x

\int \frac{1}{x ^{2} + 20 x} d x

\frac{\partial}{\partial x} (2 x - 3 y^{2} + z^{2})