What is the integral from 0 to 2 of 2pi(5-x)(8-x^3) ?

The integral from 0 to 2 of 2pi(5-x)(8-x^3) is (504pi)/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2 of 2pi(5-x)(8-x^3)

\int_{0}^{2} 2 π (5 - x) (8 - x^{3}) d x

\frac{504 π}{5}

Solution steps

\int_{0}^{2} 2 π (5 - x) (8 - x^{3}) d x

Take the constant out:

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

= 2 π \cdot \int_{0}^{2} (5 - x) (8 - x^{3}) d x

Expand

(5 - x) (8 - x^{3}) : 40 - 5 x^{3} - 8 x + x^{4}

= 2 π \cdot \int_{0}^{2} 40 - 5 x^{3} - 8 x + 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

= 2 π (\int_{0}^{2} 40 d x - \int_{0}^{2} 5 x^{3} d x - \int_{0}^{2} 8 x d x + \int_{0}^{2} x^{4} d x)

\int_{0}^{2} 40 d x = 80

\int_{0}^{2} 5 x^{3} d x = 20

\int_{0}^{2} 8 x d x = 16

\int_{0}^{2} x^{4} d x = \frac{32}{5}

= 2 π (80 - 20 - 16 + \frac{32}{5})

Simplify

2 π (80 - 20 - 16 + \frac{32}{5}) : \frac{504 π}{5}

= \frac{504 π}{5}

\int (\frac{4}{x ^{2}}) d x

\frac{x d y}{d x} + 5 y = 7 x^{2}, f (2) = 5

\frac{d}{d x} (C x e^{- x})

\int - 16 cos (4 t + π) d t

\frac{d}{d x} ((x - 2)^{x + 1})

What is the integral from 0 to 2 of 2pi(5-x)(8-x^3) ?
The integral from 0 to 2 of 2pi(5-x)(8-x^3) is (504pi)/5