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

The integral from 0 to 2 of pi(x^2+4)^2 is pi 896/15

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(x^2+4)^2

\int_{0}^{2} π (x^{2} + 4)^{2} d x

π \frac{896}{15}

Decimal

187.65780 \dots

Solution steps

\int_{0}^{2} π (x^{2} + 4)^{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}^{2} (x^{2} + 4)^{2} d x

Expand

(x^{2} + 4)^{2} : x^{4} + 8 x^{2} + 16

= π \cdot \int_{0}^{2} x^{4} + 8 x^{2} + 16 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}^{2} x^{4} d x + \int_{0}^{2} 8 x^{2} d x + \int_{0}^{2} 16 d x)

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

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

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

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

Join

\frac{32}{5} + \frac{64}{3} + 32 : \frac{896}{15}

= π \frac{896}{15}

\int_{0}^{1} \frac{1}{1 + t} d t

\int_{0}^{0.19} \frac{sin ( x )}{x} d x

\int_{- x}^{x} (t^{3} + t) d t

\int_{0}^{1} 5 x 2 x + 9 d x

\int_{0}^{2} x^{2} - x + 1 d x

What is the integral from 0 to 2 of pi(x^2+4)^2 ?
The integral from 0 to 2 of pi(x^2+4)^2 is pi 896/15