What is the integral from 0 to 2 of-8x^3sqrt(x^2+4) ?

The integral from 0 to 2 of-8x^3sqrt(x^2+4) is -8*(64sqrt(2)+64)/(15)

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integral from 0 to 2 of-8x^3sqrt(x^2+4)

\int_{0}^{2} - 8 x^{3} x^{2} + 4 d x

- 8 \cdot \frac{64 2 + 64}{15}

Decimal Notation

- 82.40515 \dots

Solution steps

\int_{0}^{2} - 8 x^{3} x^{2} + 4 d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 8 \cdot \int_{0}^{2} x^{3} x^{2} + 4 d x

Apply u-substitution

: \int_{2}^{22} (u^{2} - 4) u^{2} d u

= - 8 \cdot \int_{2}^{22} (u^{2} - 4) u^{2} d u

Expand

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

= - 8 \cdot \int_{2}^{22} u^{4} - 4 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

= - 8 (\int_{2}^{22} u^{4} d u - \int_{2}^{22} 4 u^{2} d u)

\int_{2}^{22} u^{4} d u = \frac{128 2 - 32}{5}

\int_{2}^{22} 4 u^{2} d u = \frac{4 ( 16 2 - 8 )}{3}

= - 8 (\frac{128 2 - 32}{5} - \frac{4 ( 16 2 - 8 )}{3})

Join

\frac{128 2 - 32}{5} - \frac{4 ( 16 2 - 8 )}{3} : \frac{64 2 + 64}{15}

= - 8 \cdot \frac{64 2 + 64}{15}

\int_{2}^{12} \frac{1}{27 t + 36} d t

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

x^{2} - 3 x - 4

y = 7 x^{3}

\frac{\partial}{\partial x} (ln (x^{2} + y^{2} + 4))

What is the integral from 0 to 2 of-8x^3sqrt(x^2+4) ?
The integral from 0 to 2 of-8x^3sqrt(x^2+4) is -8*(64sqrt(2)+64)/(15)