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

The integral from 0 to 1 of pi(2-1/2 x)^2 is pi 37/12

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(2-1/2 x)^2

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

π \frac{37}{12}

Decimal Notation

9.68657 \dots

Solution steps

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

Simplify

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

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

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

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

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

= π (4 - 1 + \frac{1}{12})

Simplify

π (4 - 1 + \frac{1}{12}) : π \frac{37}{12}

= π \frac{37}{12}

\frac{\partial}{\partial t} (4 cos (t))

x \to 0 lim (\frac{1}{x} - \frac{1}{x})

y = \frac{5}{x ^{2} + 1}, (2, 1)

\int 7 x cos (8 x) d x

\int x cos (4 x) d x

What is the integral from 0 to 1 of pi(2-1/2 x)^2 ?
The integral from 0 to 1 of pi(2-1/2 x)^2 is pi 37/12