What is area y^2= 1/8 x,y=8x^2 ?

The answer to area y^2= 1/8 x,y=8x^2 is 1/192

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area y^2= 1/8 x,y=8x^2

area

y^{2} = \frac{1}{8} x, y = 8 x^{2}

\frac{1}{192}

Decimal Notation

0.00520 \dots

Solution steps

Isolate

y

for

y^{2} = \frac{1}{8} x : y = \frac{x}{2 2}, y = - \frac{x}{2 2}

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

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

To find the intersection points solve

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

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

\frac{x}{2 2} = 8 x^{2} : x = 0, x = \frac{1}{8}

- \frac{x}{2 2} = 8 x^{2} : x = 0

= \int_{0}^{\frac{1}{8}} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{0}^{\frac{1}{8}} \frac{x}{2 2} - 8 x^{2} d x

\int_{0}^{\frac{1}{8}} \frac{x}{2 2} - 8 x^{2} d x = \frac{1}{192}

= \frac{1}{192}

f (x) = - x^{3} - 5 x^{2} + 2, at x = - 1

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