What is the integral from 0 to 1/2 of 4/3 (1-x^3) ?

The integral from 0 to 1/2 of 4/3 (1-x^3) is 31/48

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integral from 0 to 1/2 of 4/3 (1-x^3)

\int_{0}^{\frac{1}{2}} \frac{4}{3} (1 - x^{3}) d x

\frac{31}{48}

Decimal Notation

0.64583 \dots

Solution steps

\int_{0}^{\frac{1}{2}} \frac{4}{3} (1 - x^{3}) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

= \frac{4}{3} (\int_{0}^{\frac{1}{2}} 1 d x - \int_{0}^{\frac{1}{2}} x^{3} d x)

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

\int_{0}^{\frac{1}{2}} x^{3} d x = \frac{1}{64}

= \frac{4}{3} (\frac{1}{2} - \frac{1}{64})

Simplify

\frac{4}{3} (\frac{1}{2} - \frac{1}{64}) : \frac{31}{48}

= \frac{31}{48}

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\frac{d}{d x} (\frac{- 11}{2 ( x - 2 ) ^{2}} - \frac{4}{x - 2})

\frac{2 s + 3}{s ^{2} + 17}

\int x + 1 + \frac{2}{3} x d x

What is the integral from 0 to 1/2 of 4/3 (1-x^3) ?
The integral from 0 to 1/2 of 4/3 (1-x^3) is 31/48