What is the integral from 0 to 4 of x^3sqrt(x^2+16) ?

The integral from 0 to 4 of x^3sqrt(x^2+16) is (2048sqrt(2)+2048)/(15)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 4 of x^3sqrt(x^2+16)

\int_{0}^{4} x^{3} x^{2} + 16 d x

\frac{2048 2 + 2048}{15}

Decimal

329.62062 \dots

Solution steps

\int_{0}^{4} x^{3} x^{2} + 16 d x

Apply u-substitution

= \int_{4}^{42} (u^{2} - 16) u^{2} d u

Expand

(u^{2} - 16) u^{2} : u^{4} - 16 u^{2}

= \int_{4}^{42} u^{4} - 16 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_{4}^{42} u^{4} d u - \int_{4}^{42} 16 u^{2} d u

\int_{4}^{42} u^{4} d u = \frac{4096 2 - 1024}{5}

\int_{4}^{42} 16 u^{2} d u = \frac{16 ( 128 2 - 64 )}{3}

= \frac{4096 2 - 1024}{5} - \frac{16 ( 128 2 - 64 )}{3}

Simplify

\frac{4096 2 - 1024}{5} - \frac{16 ( 128 2 - 64 )}{3} : \frac{2048 2 + 2048}{15}

= \frac{2048 2 + 2048}{15}

What is the integral from 0 to 4 of x^3sqrt(x^2+16) ?
The integral from 0 to 4 of x^3sqrt(x^2+16) is (2048sqrt(2)+2048)/(15)