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)

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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

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

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}

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)