What is the integral from 1 to 4 of (t^2-t-6) ?

The integral from 1 to 4 of (t^2-t-6) is -9/2

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integral from 1 to 4 of (t^2-t-6)

\int_{1}^{4} (t^{2} - t - 6) d t

- \frac{9}{2}

Decimal

- 4.5

Solution steps

\int_{1}^{4} 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}^{4} t^{2} d t - \int_{1}^{4} t d t - \int_{1}^{4} 6 d t

\int_{1}^{4} t^{2} d t = 21

\int_{1}^{4} t d t = \frac{15}{2}

\int_{1}^{4} 6 d t = 18

= 21 - \frac{15}{2} - 18

Simplify

21 - \frac{15}{2} - 18 : - \frac{9}{2}

= - \frac{9}{2}

What is the integral from 1 to 4 of (t^2-t-6) ?
The integral from 1 to 4 of (t^2-t-6) is -9/2