What is the integral from 1 to 3 of (-t^2+t+6) ?

The integral from 1 to 3 of (-t^2+t+6) is 22/3

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integral from 1 to 3 of (-t^2+t+6)

\int_{1}^{3} (- t^{2} + t + 6) d t

\frac{22}{3}

Decimal

7.33333 \dots

Solution steps

\int_{1}^{3} - t^{2} + t + 6 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} t d t + \int_{1}^{3} 6 d t

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

\int_{1}^{3} t d t = 4

\int_{1}^{3} 6 d t = 12

= - \frac{26}{3} + 4 + 12

Simplify

- \frac{26}{3} + 4 + 12 : \frac{22}{3}

= \frac{22}{3}

\int_{- 1}^{2} 12 x cos (3 x^{2}) d x

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What is the integral from 1 to 3 of (-t^2+t+6) ?
The integral from 1 to 3 of (-t^2+t+6) is 22/3