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

The integral from-1 to 4 of-x^2+4x-8 is -95/3

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

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

- \frac{95}{3}

Decimal

- 31.66666 \dots

Solution steps

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

\int_{- 1}^{4} x^{2} d x = \frac{65}{3}

\int_{- 1}^{4} 4 x d x = 30

\int_{- 1}^{4} 8 d x = 40

= - \frac{65}{3} + 30 - 40

Simplify

- \frac{65}{3} + 30 - 40 : - \frac{95}{3}

= - \frac{95}{3}

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What is the integral from-1 to 4 of-x^2+4x-8 ?
The integral from-1 to 4 of-x^2+4x-8 is -95/3