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

The integral from 1/2 to 1 of 4/9 x(5-x^2) is 35/48

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{35}{48}

Decimal Notation

0.72916 \dots

Solution steps

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

Expand

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

= \frac{4}{9} \cdot \int_{\frac{1}{2}}^{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_{\frac{1}{2}}^{1} 5 x d x - \int_{\frac{1}{2}}^{1} x^{3} d x)

\int_{\frac{1}{2}}^{1} 5 x d x = \frac{15}{8}

\int_{\frac{1}{2}}^{1} x^{3} d x = \frac{15}{64}

= \frac{4}{9} (\frac{15}{8} - \frac{15}{64})

Simplify

\frac{4}{9} (\frac{15}{8} - \frac{15}{64}) : \frac{35}{48}

= \frac{35}{48}

\int_{1}^{t} 39 x^{- 3} d x

\frac{d}{d x} ((x) arctan (x))

\int_{36}^{x} 1 + 9 x d x

\frac{t - t}{t ^{\frac{1}{7}}}

\frac{3 x ^{2} + 1}{2 x}

What is the integral from 1/2 to 1 of 4/9 x(5-x^2) ?
The integral from 1/2 to 1 of 4/9 x(5-x^2) is 35/48