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

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

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

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

1

Solution steps

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

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

\int_{0}^{1} x^{3} d x = \frac{1}{4}

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

Simplify

\frac{4}{3} (1 - \frac{1}{4}) : 1

= 1

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