What is the integral from 0 to 1 of 2pi(2-y)(4y^2) ?

The integral from 0 to 1 of 2pi(2-y)(4y^2) is (10pi)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of 2pi(2-y)(4y^2)

\int_{0}^{1} 2 π (2 - y) (4 y^{2}) d y

\frac{10 π}{3}

Decimal Notation

10.47197 \dots

Solution steps

\int_{0}^{1} 2 π (2 - y) \cdot 4 y^{2} d y

Take the constant out:

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

= 2 π 4 \cdot \int_{0}^{1} (2 - y) y^{2} d y

Simplify

= 8 π \cdot \int_{0}^{1} (2 - y) y^{2} d y

Expand

(2 - y) y^{2} : 2 y^{2} - y^{3}

= 8 π \cdot \int_{0}^{1} 2 y^{2} - y^{3} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 8 π (\int_{0}^{1} 2 y^{2} d y - \int_{0}^{1} y^{3} d y)

\int_{0}^{1} 2 y^{2} d y = \frac{2}{3}

\int_{0}^{1} y^{3} d y = \frac{1}{4}

= 8 π (\frac{2}{3} - \frac{1}{4})

Simplify

8 π (\frac{2}{3} - \frac{1}{4}) : \frac{10 π}{3}

= \frac{10 π}{3}

\int - sin (d) x d x

\int (x + 1) e^{3 x^{2} + 6 x} d x

\frac{d}{d x} (A x e^{- 2 x})

\frac{d}{d x} (\frac{2 x - x ^{2}}{3 x ^{2} + 2})

What is the integral from 0 to 1 of 2pi(2-y)(4y^2) ?
The integral from 0 to 1 of 2pi(2-y)(4y^2) is (10pi)/3