What is the integral from 1 to 2 of (2x^2)-(2x+6) ?

The integral from 1 to 2 of (2x^2)-(2x+6) is -13/3

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integral from 1 to 2 of (2x^2)-(2x+6)

\int_{1}^{2} (2 x^{2}) - (2 x + 6) d x

- \frac{13}{3}

Decimal Notation

- 4.33333 \dots

Solution steps

\int_{1}^{2} 2 x^{2} - (2 x + 6) d x

Simplify

= \int_{1}^{2} 2 x^{2} - 2 x - 6 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}^{2} 2 x^{2} d x - \int_{1}^{2} 2 x d x - \int_{1}^{2} 6 d x

\int_{1}^{2} 2 x^{2} d x = \frac{14}{3}

\int_{1}^{2} 2 x d x = 3

\int_{1}^{2} 6 d x = 6

= \frac{14}{3} - 3 - 6

Simplify

\frac{14}{3} - 3 - 6 : - \frac{13}{3}

= - \frac{13}{3}

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What is the integral from 1 to 2 of (2x^2)-(2x+6) ?
The integral from 1 to 2 of (2x^2)-(2x+6) is -13/3