What is the integral from 1 to 8 of x^{1/3}-x^{-1/3} ?

The integral from 1 to 8 of x^{1/3}-x^{-1/3} is 27/4

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integral from 1 to 8 of x^{1/3}-x^{-1/3}

\int_{1}^{8} x^{\frac{1}{3}} - x^{- \frac{1}{3}} d x

\frac{27}{4}

Decimal Notation

6.75

Solution steps

\int_{1}^{8} x^{\frac{1}{3}} - x^{- \frac{1}{3}} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \int_{1}^{8} x^{\frac{1}{3}} - \frac{1}{x ^{\frac{1}{3}}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{8} x^{\frac{1}{3}} d x - \int_{1}^{8} \frac{1}{x ^{\frac{1}{3}}} d x

\int_{1}^{8} x^{\frac{1}{3}} d x = \frac{45}{4}

\int_{1}^{8} \frac{1}{x ^{\frac{1}{3}}} d x = \frac{9}{2}

= \frac{45}{4} - \frac{9}{2}

Simplify

\frac{45}{4} - \frac{9}{2} : \frac{27}{4}

= \frac{27}{4}

\int_{0}^{1} y 1 - y d y

\int_{1}^{2} 3 x^{2} + 2 x d x

\int_{0}^{\frac{π}{4}} sec^{2} (y) d y

\int_{0}^{2} (k x + 1) d x

\int_{0}^{3} x^{3} + 2 d x

What is the integral from 1 to 8 of x^{1/3}-x^{-1/3} ?
The integral from 1 to 8 of x^{1/3}-x^{-1/3} is 27/4