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

The integral from 0 to 1 of 4x(1-x^2) is 1

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

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

1

Solution steps

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

Take the constant out:

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

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

Expand

x (1 - x^{2}) : x - x^{3}

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

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

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

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

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

Simplify

4 (\frac{1}{2} - \frac{1}{4}) : 1

= 1

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