What is the integral from 0 to 8 of x^3sqrt(x^2+64) ?

The integral from 0 to 8 of x^3sqrt(x^2+64) is (65536sqrt(2)+65536)/(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 8 of x^3sqrt(x^2+64)

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

\frac{65536 2 + 65536}{15}

Decimal Notation

10547.86000 \dots

Solution steps

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

Apply u-substitution

: \int_{8}^{82} (u^{2} - 64) u^{2} d u

= \int_{8}^{82} (u^{2} - 64) u^{2} d u

Expand

(u^{2} - 64) u^{2} : u^{4} - 64 u^{2}

= \int_{8}^{82} u^{4} - 64 u^{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{8}^{82} u^{4} d u - \int_{8}^{82} 64 u^{2} d u

\int_{8}^{82} u^{4} d u = \frac{131072 2 - 32768}{5}

\int_{8}^{82} 64 u^{2} d u = \frac{64 ( 1024 2 - 512 )}{3}

= \frac{131072 2 - 32768}{5} - \frac{64 ( 1024 2 - 512 )}{3}

Simplify

\frac{131072 2 - 32768}{5} - \frac{64 ( 1024 2 - 512 )}{3} : \frac{65536 2 + 65536}{15}

= \frac{65536 2 + 65536}{15}

\frac{\partial}{\partial x} (e^{- 20})

\frac{d}{d x} (2 x^{6} - 9 x^{5} + 15 x^{4} - 10 x^{3} + 3 x - 1)

\frac{d}{d x} (\frac{x ^{2} + x - 12}{x - 4})

x \to 0 lim (\frac{2 x ^{2} + x}{x})

y = \frac{x ^{2}}{5 - 8 x}

What is the integral from 0 to 8 of x^3sqrt(x^2+64) ?
The integral from 0 to 8 of x^3sqrt(x^2+64) is (65536sqrt(2)+65536)/(15)