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

The integral from 0 to 1 of pi(4-x^2)^2 is pi 203/15

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of pi(4-x^2)^2

\int_{0}^{1} π (4 - x^{2})^{2} d x

π \frac{203}{15}

Decimal Notation

42.51622 \dots

Solution steps

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

Expand

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

= π \cdot \int_{0}^{1} 16 - 8 x^{2} + x^{4} 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} 8 x^{2} d x + \int_{0}^{1} x^{4} d x)

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

\int_{0}^{1} 8 x^{2} d x = \frac{8}{3}

\int_{0}^{1} x^{4} d x = \frac{1}{5}

= π (16 - \frac{8}{3} + \frac{1}{5})

Join

16 - \frac{8}{3} + \frac{1}{5} : \frac{203}{15}

= π \frac{203}{15}

\int_{1}^{x} r t^{r - 1} d t

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

\int_{- π}^{π} x \cdot cos (n x) d x

\int_{s}^{\infty} \frac{1}{s ^{2} + 1} d s

\int_{0}^{1} 11 ln (x) d x

What is the integral from 0 to 1 of pi(4-x^2)^2 ?
The integral from 0 to 1 of pi(4-x^2)^2 is pi 203/15