What is the integral from 0 to 2 of (2x^2-6x)/2 ?

The integral from 0 to 2 of (2x^2-6x)/2 is -10/3

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integral from 0 to 2 of (2x^2-6x)/2

\int_{0}^{2} \frac{2 x ^{2} - 6 x}{2} d x

- \frac{10}{3}

Decimal Notation

- 3.33333 \dots

Solution steps

\int_{0}^{2} \frac{2 x ^{2} - 6 x}{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{2} 2 x^{2} - 6 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

= \frac{1}{2} (\int_{0}^{2} 2 x^{2} d x - \int_{0}^{2} 6 x d x)

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

\int_{0}^{2} 6 x d x = 12

= \frac{1}{2} (\frac{16}{3} - 12)

Simplify

\frac{1}{2} (\frac{16}{3} - 12) : - \frac{10}{3}

= - \frac{10}{3}

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What is the integral from 0 to 2 of (2x^2-6x)/2 ?
The integral from 0 to 2 of (2x^2-6x)/2 is -10/3