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)

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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 Notation

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

= \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}

\frac{d}{d x} (3 x^{2} + 2 x + 1)

t^{2} y^{^{'}} + t y = 5

x sin (3 x)

\frac{d}{d x} (\frac{1}{e ^{x} + 1})

x \cdot ln (x)

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)