What is the integral from 0.5 to 1 of 4/9 x(5-x^2) ?

The integral from 0.5 to 1 of 4/9 x(5-x^2) is 0.72916…

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integral from 0.5 to 1 of 4/9 x(5-x^2)

\int_{0.5}^{1} \frac{4}{9} x (5 - x^{2}) d x

0.72916 \dots

Solution steps

\int_{0.5}^{1} \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_{0.5}^{1} x (5 - x^{2}) d x

Expand

x (5 - x^{2}) : 5 x - x^{3}

= \frac{4}{9} \cdot \int_{0.5}^{1} 5 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

= \frac{4}{9} (\int_{0.5}^{1} 5 x d x - \int_{0.5}^{1} x^{3} d x)

\int_{0.5}^{1} 5 x d x = 1.875

\int_{0.5}^{1} x^{3} d x = 0.234375

= \frac{4}{9} (1.875 - 0.234375)

Simplify

\frac{4}{9} (1.875 - 0.234375) : 0.72916 \dots

= 0.72916 \dots

x \to 0 - lim (- 2 x^{2} + 2 x)

\int \frac{3}{x ( x - 3 ) ^{2}} d x

\int sec^{2} (y) d y

n = 1 \sum \infty (\frac{8}{9})^{n}

x \to 0 lim (\frac{7 x + x ^{2}}{x})

What is the integral from 0.5 to 1 of 4/9 x(5-x^2) ?
The integral from 0.5 to 1 of 4/9 x(5-x^2) is 0.72916…