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

The integral from 0 to 1 of pi(4-1/2 x)^2 is pi 169/12

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 pi(4-1/2 x)^2

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

π \frac{169}{12}

Decimal

44.24409 \dots

Solution steps

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

Take the constant out:

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

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

Simplify

(4 - \frac{1}{2} x)^{2} : 16 - 4 x + \frac{1}{4} x^{2}

= π \cdot \int_{0}^{1} 16 - 4 x + \frac{1}{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

= π (\int_{0}^{1} 16 d x - \int_{0}^{1} 4 x d x + \int_{0}^{1} \frac{1}{4} x^{2} d x)

\int_{0}^{1} 16 d x = 16

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

\int_{0}^{1} \frac{1}{4} x^{2} d x = \frac{1}{12}

= π (16 - 2 + \frac{1}{12})

Simplify

π (16 - 2 + \frac{1}{12}) : π \frac{169}{12}

= π \frac{169}{12}

(2 x - 9) (x^{2} + 6), at x = 3

\int_{0}^{6} 1

\int x - 7 + 8 x - x^{2} d x

x \to 7 lim (6 x^{- 1})

x \to 2 lim (3 x^{2} - 2 x + 4)

What is the integral from 0 to 1 of pi(4-1/2 x)^2 ?
The integral from 0 to 1 of pi(4-1/2 x)^2 is pi 169/12