What is the integral from-1 to 0 of t^{1/3}-t^{2/3} ?

The integral from-1 to 0 of t^{1/3}-t^{2/3} is -27/20

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integral from-1 to 0 of t^{1/3}-t^{2/3}

\int_{- 1}^{0} t^{\frac{1}{3}} - t^{\frac{2}{3}} d t

- \frac{27}{20}

Decimal Notation

- 1.35

Solution steps

\int_{- 1}^{0} t^{\frac{1}{3}} - t^{\frac{2}{3}} d t

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}^{0} t^{\frac{1}{3}} d t - \int_{- 1}^{0} t^{\frac{2}{3}} d t

\int_{- 1}^{0} t^{\frac{1}{3}} d t = \frac{3 ( - 1 ) ^{\frac{1}{3}}}{4}

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

= \frac{3 ( - 1 ) ^{\frac{1}{3}}}{4} - \frac{3 ( - 1 ) ^{\frac{2}{3}}}{5}

Simplify

\frac{3 ( - 1 ) ^{\frac{1}{3}}}{4} - \frac{3 ( - 1 ) ^{\frac{2}{3}}}{5} : - \frac{27}{20}

= - \frac{27}{20}

f (x) = - \frac{1}{4} x^{2}, (- 2, - 1)

\frac{1}{( s - 1 ) ^{3}}

\frac{d}{d x} ((8 x + e^{- x^{2}})^{3})

f (x) = \frac{6}{x ^{3}}

\frac{d y}{d x} = \frac{11 x}{y}

What is the integral from-1 to 0 of t^{1/3}-t^{2/3} ?
The integral from-1 to 0 of t^{1/3}-t^{2/3} is -27/20