What is the integral from-3 to 3 of (-5x^2-6x)^3 ?

The integral from-3 to 3 of (-5x^2-6x)^3 is -914166/7

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integral from-3 to 3 of (-5x^2-6x)^3

\int_{- 3}^{3} (- 5 x^{2} - 6 x)^{3} d x

- \frac{914166}{7}

Decimal

- 130595.14285 \dots

Solution steps

\int_{- 3}^{3} (- 5 x^{2} - 6 x)^{3} d x

Expand

(- 5 x^{2} - 6 x)^{3} : - 125 x^{6} - 450 x^{5} - 540 x^{4} - 216 x^{3}

= \int_{- 3}^{3} - 125 x^{6} - 450 x^{5} - 540 x^{4} - 216 x^{3} 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

= - \int_{- 3}^{3} 125 x^{6} d x - \int_{- 3}^{3} 450 x^{5} d x - \int_{- 3}^{3} 540 x^{4} d x - \int_{- 3}^{3} 216 x^{3} d x

\int_{- 3}^{3} 125 x^{6} d x = \frac{546750}{7}

\int_{- 3}^{3} 450 x^{5} d x = 0

\int_{- 3}^{3} 540 x^{4} d x = 52488

\int_{- 3}^{3} 216 x^{3} d x = 0

= - \frac{546750}{7} - 0 - 52488 - 0

Simplify

- \frac{546750}{7} - 0 - 52488 - 0 : - \frac{914166}{7}

= - \frac{914166}{7}

\int_{3}^{\infty} x e^{- x^{2}} d x

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\int_{0}^{2} x 4 x^{2} + 1 d x

\int_{3}^{5} 6 + x^{2} d x

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

What is the integral from-3 to 3 of (-5x^2-6x)^3 ?
The integral from-3 to 3 of (-5x^2-6x)^3 is -914166/7