What is the integral from 1 to 3 of t^2+3t-18 ?

The integral from 1 to 3 of t^2+3t-18 is -46/3

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integral from 1 to 3 of t^2+3t-18

\int_{1}^{3} t^{2} + 3 t - 18 d t

- \frac{46}{3}

Decimal

- 15.33333 \dots

Solution steps

\int_{1}^{3} t^{2} + 3 t - 18 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}^{3} t^{2} d t + \int_{1}^{3} 3 t d t - \int_{1}^{3} 18 d t

\int_{1}^{3} t^{2} d t = \frac{26}{3}

\int_{1}^{3} 3 t d t = 12

\int_{1}^{3} 18 d t = 36

= \frac{26}{3} + 12 - 36

Simplify

\frac{26}{3} + 12 - 36 : - \frac{46}{3}

= - \frac{46}{3}

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What is the integral from 1 to 3 of t^2+3t-18 ?
The integral from 1 to 3 of t^2+3t-18 is -46/3