What is the integral from-1 to 2 of (-x^4-x^2+6x+8) ?

The integral from-1 to 2 of (-x^4-x^2+6x+8) is 117/5

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integral from-1 to 2 of (-x^4-x^2+6x+8)

\int_{- 1}^{2} (- x^{4} - x^{2} + 6 x + 8) d x

\frac{117}{5}

Decimal Notation

23.4

Solution steps

\int_{- 1}^{2} - x^{4} - x^{2} + 6 x + 8 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} x^{4} d x - \int_{- 1}^{2} x^{2} d x + \int_{- 1}^{2} 6 x d x + \int_{- 1}^{2} 8 d x

\int_{- 1}^{2} x^{4} d x = \frac{33}{5}

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

\int_{- 1}^{2} 6 x d x = 9

\int_{- 1}^{2} 8 d x = 24

= - \frac{33}{5} - 3 + 9 + 24

Simplify

- \frac{33}{5} - 3 + 9 + 24 : \frac{117}{5}

= \frac{117}{5}

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What is the integral from-1 to 2 of (-x^4-x^2+6x+8) ?
The integral from-1 to 2 of (-x^4-x^2+6x+8) is 117/5