What is the integral from 0 to 3 of pi(9x^2-x^4) ?

The integral from 0 to 3 of pi(9x^2-x^4) is pi 162/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 3 of pi(9x^2-x^4)

\int_{0}^{3} π (9 x^{2} - x^{4}) d x

π \frac{162}{5}

Decimal

101.78760 \dots

Solution steps

\int_{0}^{3} π (9 x^{2} - x^{4}) d x

Take the constant out:

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

= π \cdot \int_{0}^{3} 9 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

= π (\int_{0}^{3} 9 x^{2} d x - \int_{0}^{3} x^{4} d x)

\int_{0}^{3} 9 x^{2} d x = 81

\int_{0}^{3} x^{4} d x = \frac{243}{5}

= π (81 - \frac{243}{5})

Join

81 - \frac{243}{5} : \frac{162}{5}

= π \frac{162}{5}

What is the integral from 0 to 3 of pi(9x^2-x^4) ?
The integral from 0 to 3 of pi(9x^2-x^4) is pi 162/5