What is the integral from 0 to 1 of x(4/9 x(5-x^2)) ?

The integral from 0 to 1 of x(4/9 x(5-x^2)) is 88/135

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integral from 0 to 1 of x(4/9 x(5-x^2))

\int_{0}^{1} x (\frac{4}{9} x (5 - x^{2})) d x

\frac{88}{135}

Decimal Notation

0.65185 \dots

Solution steps

\int_{0}^{1} x \frac{4}{9} x (5 - x^{2}) d x

Take the constant out:

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

= \frac{4}{9} \cdot \int_{0}^{1} xx (5 - x^{2}) d x

Simplify

xx (5 - x^{2}) : x^{2} (5 - x^{2})

= \frac{4}{9} \cdot \int_{0}^{1} x^{2} (5 - x^{2}) d x

Expand

x^{2} (5 - x^{2}) : 5 x^{2} - x^{4}

= \frac{4}{9} \cdot \int_{0}^{1} 5 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

= \frac{4}{9} (\int_{0}^{1} 5 x^{2} d x - \int_{0}^{1} x^{4} d x)

\int_{0}^{1} 5 x^{2} d x = \frac{5}{3}

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

= \frac{4}{9} (\frac{5}{3} - \frac{1}{5})

Simplify

\frac{4}{9} (\frac{5}{3} - \frac{1}{5}) : \frac{88}{135}

= \frac{88}{135}

x \to 3 lim (5 + x)

\frac{\partial}{\partial x} (3 x^{2} - 2 y^{2})

\frac{\partial}{\partial x} (ln (6 x^{2} + 7 y^{2} + 3))

\frac{\partial}{\partial x} (x^{4} y^{6} + 4 x^{7} y)

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = x - 3 e^{x}

What is the integral from 0 to 1 of x(4/9 x(5-x^2)) ?
The integral from 0 to 1 of x(4/9 x(5-x^2)) is 88/135