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

The integral from 0 to 2 of 2pix(3x(x-2)^2) is (32pi)/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 2pix(3x(x-2)^2)

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

\frac{32 π}{5}

Decimal Notation

20.10619 \dots

Solution steps

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

Take the constant out:

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

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

Simplify

= 6 π \cdot \int_{0}^{2} xx (x - 2)^{2} d x

Expand

xx (x - 2)^{2} : x^{4} - 4 x^{3} + 4 x^{2}

= 6 π \cdot \int_{0}^{2} x^{4} - 4 x^{3} + 4 x^{2} 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

= 6 π (\int_{0}^{2} x^{4} d x - \int_{0}^{2} 4 x^{3} d x + \int_{0}^{2} 4 x^{2} d x)

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

\int_{0}^{2} 4 x^{3} d x = 16

\int_{0}^{2} 4 x^{2} d x = \frac{32}{3}

= 6 π (\frac{32}{5} - 16 + \frac{32}{3})

Simplify

6 π (\frac{32}{5} - 16 + \frac{32}{3}) : \frac{32 π}{5}

= \frac{32 π}{5}

n = 1 \sum \infty \frac{2 n}{3 n - 1}

\frac{d}{d x} (\frac{3 a}{4 x ^{5}})

\frac{x ^{2} + 4 x}{( 2 + x ) ^{2}}

ln (1 - 2 x)

\int (2 x + 4)^{3} d x

What is the integral from 0 to 2 of 2pix(3x(x-2)^2) ?
The integral from 0 to 2 of 2pix(3x(x-2)^2) is (32pi)/5