What is the integral from 0 to 2 of-6x^3sqrt(x^2+4) ?

The integral from 0 to 2 of-6x^3sqrt(x^2+4) is -(2(64sqrt(2)+64))/5

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integral from 0 to 2 of-6x^3sqrt(x^2+4)

\int_{0}^{2} - 6 x^{3} x^{2} + 4 d x

- \frac{2 ( 64 2 + 64 )}{5}

Decimal

- 61.80386 \dots

Solution steps

\int_{0}^{2} - 6 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

= - 6 \cdot \int_{0}^{2} x^{3} x^{2} + 4 d x

Apply u-substitution

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

= - 6 \cdot \int_{2}^{22} (u^{2} - 4) u^{2} d u

Expand

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

= - 6 \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

= - 6 (\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}

= - 6 (\frac{128 2 - 32}{5} - \frac{4 ( 16 2 - 8 )}{3})

Simplify

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

= - \frac{2 ( 64 2 + 64 )}{5}

\frac{d}{d x} (1 + x + 5 x^{2})

\int_{- 2}^{1} 2 - x^{2} - x d x

(x^{9} - 1)^{3}

y = 3 x, y = 2 x - 18, y = 0

\frac{x ^{2} - 6 x + 1}{x ^{2} + 7}

What is the integral from 0 to 2 of-6x^3sqrt(x^2+4) ?
The integral from 0 to 2 of-6x^3sqrt(x^2+4) is -(2(64sqrt(2)+64))/5