What is the integral from 1 to 6 of t^2-2t-8 ?

The integral from 1 to 6 of t^2-2t-8 is -10/3

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integral from 1 to 6 of t^2-2t-8

\int_{1}^{6} t^{2} - 2 t - 8 d t

- \frac{10}{3}

Solution steps

\int_{1}^{6} t^{2} - 2 t - 8 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}^{6} t^{2} d t - \int_{1}^{6} 2 t d t - \int_{1}^{6} 8 d t

\int_{1}^{6} t^{2} d t = \frac{215}{3}

\int_{1}^{6} 2 t d t = 35

\int_{1}^{6} 8 d t = 40

= \frac{215}{3} - 35 - 40

Simplify

\frac{215}{3} - 35 - 40 : - \frac{10}{3}

= - \frac{10}{3}

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What is the integral from 1 to 6 of t^2-2t-8 ?
The integral from 1 to 6 of t^2-2t-8 is -10/3