What is the integral from 1 to 8 of (x^2-4x-12) ?

The integral from 1 to 8 of (x^2-4x-12) is -119/3

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integral from 1 to 8 of (x^2-4x-12)

\int_{1}^{8} (x^{2} - 4 x - 12) d x

- \frac{119}{3}

Decimal Notation

- 39.66666 \dots

Solution steps

\int_{1}^{8} x^{2} - 4 x - 12 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}^{8} x^{2} d x - \int_{1}^{8} 4 x d x - \int_{1}^{8} 12 d x

\int_{1}^{8} x^{2} d x = \frac{511}{3}

\int_{1}^{8} 4 x d x = 126

\int_{1}^{8} 12 d x = 84

= \frac{511}{3} - 126 - 84

Simplify

\frac{511}{3} - 126 - 84 : - \frac{119}{3}

= - \frac{119}{3}

\int_{0}^{1} 2 x^{2} + 1 d x

\int_{2}^{4} 16 - r^{2} r d r

\int_{1}^{6} (\frac{9}{x} - e^{- x}) d x

\int_{0}^{3} (x^{3} - 4 x) d x

\int_{0}^{3} \frac{3}{x + 4} d x

What is the integral from 1 to 8 of (x^2-4x-12) ?
The integral from 1 to 8 of (x^2-4x-12) is -119/3