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

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

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

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

\frac{8}{3}

Decimal Notation

2.66666 \dots

Solution steps

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

\int_{- 2}^{0} 4 x^{2} d x = \frac{32}{3}

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

= \frac{32}{3} - 8

Simplify

\frac{32}{3} - 8 : \frac{8}{3}

= \frac{8}{3}

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