What is the integral from 0 to 3 of x(x^2-1)^3 ?

The integral from 0 to 3 of x(x^2-1)^3 is 4095/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 3 of x(x^2-1)^3

\int_{0}^{3} x (x^{2} - 1)^{3} d x

\frac{4095}{8}

Decimal

511.875

Solution steps

\int_{0}^{3} x (x^{2} - 1)^{3} d x

Expand

x (x^{2} - 1)^{3} : x^{7} - 3 x^{5} + 3 x^{3} - x

= \int_{0}^{3} x^{7} - 3 x^{5} + 3 x^{3} - x 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

= \int_{0}^{3} x^{7} d x - \int_{0}^{3} 3 x^{5} d x + \int_{0}^{3} 3 x^{3} d x - \int_{0}^{3} x d x

\int_{0}^{3} x^{7} d x = \frac{6561}{8}

\int_{0}^{3} 3 x^{5} d x = \frac{729}{2}

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

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

= \frac{6561}{8} - \frac{729}{2} + \frac{243}{4} - \frac{9}{2}

Simplify

\frac{6561}{8} - \frac{729}{2} + \frac{243}{4} - \frac{9}{2} : \frac{4095}{8}

= \frac{4095}{8}

\int_{0}^{\frac{1}{2}} 4 x d x

\int_{0}^{1} 1 x + 3 d x

\int_{0}^{\frac{π}{2}} 2 sin (\frac{x}{2}) d x

\int_{0}^{π} x sin (x) cos (n x) d x

\int_{0}^{l n (4)} 3 x^{2} e^{3 x} d x

What is the integral from 0 to 3 of x(x^2-1)^3 ?
The integral from 0 to 3 of x(x^2-1)^3 is 4095/8