What is area y=4x^2,y^2= 1/4 x ?

The answer to area y=4x^2,y^2= 1/4 x is 1/48

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area y=4x^2,y^2= 1/4 x

area

y = 4 x^{2}, y^{2} = \frac{1}{4} x

\frac{1}{48}

Decimal Notation

0.02083 \dots

Solution steps

Isolate

y

for

y^{2} = \frac{1}{4} x : y = \frac{x}{2}, y = - \frac{x}{2}

f_{1} (x) = 4 x^{2}

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

To find the intersection points solve

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

4 x^{2} = \frac{x}{2} : x = 0, x = \frac{1}{4}

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

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

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

= \int_{0}^{\frac{1}{4}} 4 x^{2} - \frac{x}{2} d x

\int_{0}^{\frac{1}{4}} 4 x^{2} - \frac{x}{2} d x = \frac{1}{48}

= \frac{1}{48}

y^{^{''}} - x^{2} + y^{^{'}} = 0

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

f (x) = 7^{x}

\frac{2}{3} t^{3} \cdot e^{- 8 t}

\frac{d}{d x} (\frac{4}{x ^{2}} - \frac{2}{x})