What is the integral from-1 to 3 of (2x^2-4x) ?

The integral from-1 to 3 of (2x^2-4x) is 8/3

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integral from-1 to 3 of (2x^2-4x)

\int_{- 1}^{3} (2 x^{2} - 4 x) d x

\frac{8}{3}

Decimal

2.66666 \dots

Solution steps

\int_{- 1}^{3} 2 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_{- 1}^{3} 2 x^{2} d x - \int_{- 1}^{3} 4 x d x

\int_{- 1}^{3} 2 x^{2} d x = \frac{56}{3}

\int_{- 1}^{3} 4 x d x = 16

= \frac{56}{3} - 16

Simplify

\frac{56}{3} - 16 : \frac{8}{3}

= \frac{8}{3}

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What is the integral from-1 to 3 of (2x^2-4x) ?
The integral from-1 to 3 of (2x^2-4x) is 8/3