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

The integral from-1 to 6 of (2x+6)-(x^2-3x) is 343/6

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integral from-1 to 6 of (2x+6)-(x^2-3x)

\int_{- 1}^{6} (2 x + 6) - (x^{2} - 3 x) d x

\frac{343}{6}

Decimal Notation

57.16666 \dots

Solution steps

\int_{- 1}^{6} 2 x + 6 - (x^{2} - 3 x) d x

Simplify

= \int_{- 1}^{6} - x^{2} + 5 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}^{6} x^{2} d x + \int_{- 1}^{6} 5 x d x + \int_{- 1}^{6} 6 d x

\int_{- 1}^{6} x^{2} d x = \frac{217}{3}

\int_{- 1}^{6} 5 x d x = \frac{175}{2}

\int_{- 1}^{6} 6 d x = 42

= - \frac{217}{3} + \frac{175}{2} + 42

Simplify

- \frac{217}{3} + \frac{175}{2} + 42 : \frac{343}{6}

= \frac{343}{6}

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What is the integral from-1 to 6 of (2x+6)-(x^2-3x) ?
The integral from-1 to 6 of (2x+6)-(x^2-3x) is 343/6