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 Notation

- 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_{4}^{6} \frac{1}{x x ^{2} - 4} d x

\int_{4}^{7} (2 x^{2} - x) d x

\int_{0}^{6} 80 x e^{\frac{- x}{10}} d x

\int_{- 7}^{0} \frac{e ^{\frac{1}{x}}}{x ^{2}} d x

\int_{0}^{1} \frac{1}{2} e^{- \frac{1}{2} x} d x

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