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

The integral from 0 to 2 of pi(4-y^2)^2 is pi 256/15

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 pi(4-y^2)^2

\int_{0}^{2} π (4 - y^{2})^{2} d y

π \frac{256}{15}

Decimal

53.61651 \dots

Solution steps

\int_{0}^{2} π (4 - y^{2})^{2} d y

Take the constant out:

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

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

Expand

(4 - y^{2})^{2} : 16 - 8 y^{2} + y^{4}

= π \cdot \int_{0}^{2} 16 - 8 y^{2} + y^{4} 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

= π (\int_{0}^{2} 16 d y - \int_{0}^{2} 8 y^{2} d y + \int_{0}^{2} y^{4} d y)

\int_{0}^{2} 16 d y = 32

\int_{0}^{2} 8 y^{2} d y = \frac{64}{3}

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

= π (32 - \frac{64}{3} + \frac{32}{5})

Join

32 - \frac{64}{3} + \frac{32}{5} : \frac{256}{15}

= π \frac{256}{15}

\int_{0}^{π} 4 sin (2 x) d x

\int_{2}^{3} 95 + 1200 x^{2} - 20 x d x

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

\int_{1}^{4} e^{x} - \frac{1}{x} d x

\int_{- 1}^{1} x^{3} - 2 x^{2} - x + 2 d x

What is the integral from 0 to 2 of pi(4-y^2)^2 ?
The integral from 0 to 2 of pi(4-y^2)^2 is pi 256/15