What is the integral from 5 to-5 of 1/(|x|^{(2/3))} ?

The integral from 5 to-5 of 1/(|x|^{(2/3))} is -6*5^{1/3}

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integral from 5 to-5 of 1/(|x|^{(2/3))}

\int_{5}^{- 5} \frac{1}{∣ x ∣ ^{(\frac{2}{3})}} d x

- 6 \cdot 5^{\frac{1}{3}}

Decimal

- 10.25985 \dots

Solution steps

\int_{5}^{- 5} \frac{1}{∣ x ∣ ^{\frac{2}{3}}} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 5}^{5} \frac{1}{∣ x ∣ ^{\frac{2}{3}}} d x

Eliminate Absolutes

= - (\int_{- 5}^{0} \frac{1}{( - x ) ^{\frac{2}{3}}} d x + \int_{0}^{5} \frac{1}{x ^{\frac{2}{3}}} d x)

\int_{- 5}^{0} \frac{1}{( - x ) ^{\frac{2}{3}}} d x = - 3 (- 1)^{\frac{1}{3}} \cdot 5^{\frac{1}{3}}

\int_{0}^{5} \frac{1}{x ^{\frac{2}{3}}} d x = 3 \cdot 5^{\frac{1}{3}}

= - (- 3 (- 1)^{\frac{1}{3}} \cdot 5^{\frac{1}{3}} + 3 \cdot 5^{\frac{1}{3}})

Simplify

- (- 3 (- 1)^{\frac{1}{3}} \cdot 5^{\frac{1}{3}} + 3 \cdot 5^{\frac{1}{3}}) : - 6 \cdot 5^{\frac{1}{3}}

= - 6 \cdot 5^{\frac{1}{3}}

n = 0 \sum \infty \frac{ln ( n )}{n !}

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = 3, y^{^{'}} (0) = 2

x^{2} + 3 x

\frac{d}{d x} (5 x lo g_{4} (2 x - 1))

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

What is the integral from 5 to-5 of 1/(|x|^{(2/3))} ?
The integral from 5 to-5 of 1/(|x|^{(2/3))} is -6*5^{1/3}