What is area 1/(14x^2),x, x/4 ?

The answer to area 1/(14x^2),x, x/4 is (6\sqrt[3]{14}-3*2^{2/3}\sqrt[3]{7})/(56)

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area 1/(14x^2),x, x/4

area

\frac{1}{14 x ^{2}}, x, \frac{x}{4}

\frac{6 3 14 - 3 \cdot 2 ^{\frac{2}{3}} 3 7}{56}

Decimal

0.09555 \dots

Solution steps

To find the intersection points solve

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

\frac{1}{14 x ^{2}} = x : x = \frac{1}{3 14}

\frac{1}{14 x ^{2}} = \frac{x}{4} : x = 3 \frac{2}{7}

x = \frac{x}{4} : x = 0

= \int_{0}^{\frac{1}{3 14}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{1}{3 14}}^{3 \frac{2}{7}} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{0}^{\frac{1}{3 14}} x - \frac{x}{4} d x + \int_{\frac{1}{3 14}}^{3 \frac{2}{7}} \frac{1}{14 x ^{2}} - \frac{x}{4} d x

\int_{0}^{\frac{1}{3 14}} x - \frac{x}{4} d x = \frac{3}{8 \cdot 1 4 ^{\frac{2}{3}}}

\int_{\frac{1}{3 14}}^{3 \frac{2}{7}} \frac{1}{14 x ^{2}} - \frac{x}{4} d x = \frac{- 4 3 49 \cdot 2 ^{\frac{2}{3}} + 9 3 2 \cdot 7 ^{\frac{2}{3}} - 2 \cdot 1 4 ^{\frac{2}{3}}}{56 3 14 \cdot 2 ^{\frac{2}{3}}}

= \frac{3}{8 \cdot 1 4 ^{\frac{2}{3}}} + \frac{- 4 3 49 \cdot 2 ^{\frac{2}{3}} + 9 3 2 \cdot 7 ^{\frac{2}{3}} - 2 \cdot 1 4 ^{\frac{2}{3}}}{56 3 14 \cdot 2 ^{\frac{2}{3}}}

Simplify

= \frac{6 3 14 - 3 \cdot 2 ^{\frac{2}{3}} 3 7}{56}

\int x^{2} - x^{3} d x

\int_{0}^{1} \frac{1}{3 x} d x

\int_{0}^{25} (- 6 x) d x

x \to 2 lim (x^{3} - 5 x^{2} + 1)

\int_{0}^{2} (x + y) d y

What is area 1/(14x^2),x, x/4 ?
The answer to area 1/(14x^2),x, x/4 is (6\sqrt[3]{14}-3*2^{2/3}\sqrt[3]{7})/(56)