What is the integral from 1 to 8 of t^2-4t-12 ?

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

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integral from 1 to 8 of t^2-4t-12

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

- \frac{119}{3}

Solution steps

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

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

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

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

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

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

Simplify

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

= - \frac{119}{3}

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