What is the integral from-2 to 0 of 6x^2(2x^3+1)^2 ?

The integral from-2 to 0 of 6x^2(2x^3+1)^2 is 3376/3

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integral from-2 to 0 of 6x^2(2x^3+1)^2

\int_{- 2}^{0} 6 x^{2} (2 x^{3} + 1)^{2} d x

\frac{3376}{3}

Decimal

1125.33333 \dots

Solution steps

\int_{- 2}^{0} 6 x^{2} (2 x^{3} + 1)^{2} d x

Take the constant out:

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

= 6 \cdot \int_{- 2}^{0} x^{2} (2 x^{3} + 1)^{2} d x

Expand

x^{2} (2 x^{3} + 1)^{2} : 4 x^{8} + 4 x^{5} + x^{2}

= 6 \cdot \int_{- 2}^{0} 4 x^{8} + 4 x^{5} + x^{2} d x

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}^{0} 4 x^{8} d x + \int_{- 2}^{0} 4 x^{5} d x + \int_{- 2}^{0} x^{2} d x)

\int_{- 2}^{0} 4 x^{8} d x = \frac{2048}{9}

\int_{- 2}^{0} 4 x^{5} d x = - \frac{128}{3}

\int_{- 2}^{0} x^{2} d x = \frac{8}{3}

= 6 (\frac{2048}{9} - \frac{128}{3} + \frac{8}{3})

Simplify

6 (\frac{2048}{9} - \frac{128}{3} + \frac{8}{3}) : \frac{3376}{3}

= \frac{3376}{3}

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What is the integral from-2 to 0 of 6x^2(2x^3+1)^2 ?
The integral from-2 to 0 of 6x^2(2x^3+1)^2 is 3376/3