What is the integral from 0 to 0.5 of 4/3 (1-x^3) ?

The integral from 0 to 0.5 of 4/3 (1-x^3) is 0.64583…

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integral from 0 to 0.5 of 4/3 (1-x^3)

\int_{0}^{0.5} \frac{4}{3} (1 - x^{3}) d x

0.64583 \dots

Solution steps

\int_{0}^{0.5} \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}^{0.5} 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}^{0.5} 1 d x - \int_{0}^{0.5} x^{3} d x)

\int_{0}^{0.5} 1 d x = 0.5

\int_{0}^{0.5} x^{3} d x = 0.015625

= \frac{4}{3} (0.5 - 0.015625)

Simplify

\frac{4}{3} (0.5 - 0.015625) : 0.64583 \dots

= 0.64583 \dots

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What is the integral from 0 to 0.5 of 4/3 (1-x^3) ?
The integral from 0 to 0.5 of 4/3 (1-x^3) is 0.64583…