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

The integral from-2 to 3 of [(x+4)-(x^2-2)] is 125/6

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integral from-2 to 3 of [(x+4)-(x^2-2)]

\int_{- 2}^{3} [(x + 4) - (x^{2} - 2)] d x

\frac{125}{6}

Decimal

20.83333 \dots

Solution steps

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

Simplify

= \int_{- 2}^{3} - x^{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_{- 2}^{3} x^{2} d x + \int_{- 2}^{3} x d x + \int_{- 2}^{3} 6 d x

\int_{- 2}^{3} x^{2} d x = \frac{35}{3}

\int_{- 2}^{3} x d x = \frac{5}{2}

\int_{- 2}^{3} 6 d x = 30

= - \frac{35}{3} + \frac{5}{2} + 30

Simplify

- \frac{35}{3} + \frac{5}{2} + 30 : \frac{125}{6}

= \frac{125}{6}

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What is the integral from-2 to 3 of [(x+4)-(x^2-2)] ?
The integral from-2 to 3 of [(x+4)-(x^2-2)] is 125/6