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

The integral from 0 to 4 of pi(2-sqrt(x))^2 is pi 8/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 4 of pi(2-sqrt(x))^2

\int_{0}^{4} π (2 - x)^{2} d x

π \frac{8}{3}

Decimal

8.37758 \dots

Solution steps

\int_{0}^{4} π (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}^{4} (2 - x)^{2} d x

Expand

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

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

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

\int_{0}^{4} 4 x d x = \frac{64}{3}

\int_{0}^{4} x d x = 8

= π (16 - \frac{64}{3} + 8)

Simplify

π (16 - \frac{64}{3} + 8) : π \frac{8}{3}

= π \frac{8}{3}

f (x) = e^{2 x} \cdot 2

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

\frac{d x}{d y} = \frac{y}{x ^{2}}

\frac{6 - x e ^{x}}{x + e ^{x}}

\frac{\partial}{\partial y} (\frac{1}{2 x + y})

What is the integral from 0 to 4 of pi(2-sqrt(x))^2 ?
The integral from 0 to 4 of pi(2-sqrt(x))^2 is pi 8/3