What is the integral from-4 to 0 of (8x^3-19x^2) ?

The integral from-4 to 0 of (8x^3-19x^2) is -2752/3

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integral from-4 to 0 of (8x^3-19x^2)

\int_{- 4}^{0} (8 x^{3} - 19 x^{2}) d x

- \frac{2752}{3}

Decimal

- 917.33333 \dots

Solution steps

\int_{- 4}^{0} 8 x^{3} - 19 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

= \int_{- 4}^{0} 8 x^{3} d x - \int_{- 4}^{0} 19 x^{2} d x

\int_{- 4}^{0} 8 x^{3} d x = - 512

\int_{- 4}^{0} 19 x^{2} d x = \frac{1216}{3}

= - 512 - \frac{1216}{3}

Simplify

- 512 - \frac{1216}{3} : - \frac{2752}{3}

= - \frac{2752}{3}

What is the integral from-4 to 0 of (8x^3-19x^2) ?
The integral from-4 to 0 of (8x^3-19x^2) is -2752/3