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

The integral from-2 to 0 of (4x^3+x^2-2x) is -28/3

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

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

- \frac{28}{3}

Decimal

- 9.33333 \dots

Solution steps

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

\int_{- 2}^{0} 4 x^{3} d x = - 16

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

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

= - 16 + \frac{8}{3} - (- 4)

Simplify

- 16 + \frac{8}{3} - (- 4) : - \frac{28}{3}

= - \frac{28}{3}

What is the integral from-2 to 0 of (4x^3+x^2-2x) ?
The integral from-2 to 0 of (4x^3+x^2-2x) is -28/3