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

The integral from 0 to 1 of x*4(x-x^3) is 8/15

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

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

\frac{8}{15}

Decimal

0.53333 \dots

Solution steps

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

Expand

x (x - x^{3}) : x^{2} - x^{4}

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

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

\int_{0}^{1} x^{4} d x = \frac{1}{5}

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

Simplify

4 (\frac{1}{3} - \frac{1}{5}) : \frac{8}{15}

= \frac{8}{15}

What is the integral from 0 to 1 of x*4(x-x^3) ?
The integral from 0 to 1 of x*4(x-x^3) is 8/15